# 6: Factoring - Mathematics

6: Factoring - Mathematics

Factorising - Expanding Brackets

This section shows you how to factorise and includes examples, sample questions and videos.

Brackets should be expanded in the following ways:
For an expression of the form a(b + c), the expanded version is ab + ac, i.e., multiply the term outside the bracket by everything inside the bracket (e.g. 2x(x + 3) = 2x² + 6x [remember x × x is x²]).
For an expression of the form (a + b)(c + d), the expanded version is ac + ad + bc + bd, in other words everything in the first bracket should be multiplied by everything in the second.

Expand (2x + 3)(x - 1):
(2x + 3)(x - 1)
= 2x² - 2x + 3x - 3
= 2x² + x - 3

Factorising

Factorising is the reverse of expanding brackets, so it is, for example, putting 2x² + x - 3 into the form (2x + 3)(x - 1). This is an important way of solving quadratic equations.
The first step of factorising an expression is to 'take out' any common factors which the terms have. So if you were asked to factorise x² + x, since x goes into both terms, you would write x(x + 1) .

This video shows you how to solve a quadratic equation by factoring. There is no simple method of factorising a quadratic expression, but with a little practise it becomes easier. One systematic method, however, is as follows:

Factorise 12y² - 20y + 3
= 12y² - 18y - 2y + 3 [here the 20y has been split up into two numbers whose multiple is 36. 36 was chosen because this is the product of 12 and 3, the other two numbers].
The first two terms, 12y² and -18y both divide by 6y, so 'take out' this factor of 6y.
6y(2y - 3) - 2y + 3 [we can do this because 6y(2y - 3) is the same as 12y² - 18y]
Now, make the last two expressions look like the expression in the bracket:
6y(2y - 3) -1(2y - 3)
The answer is (2y - 3)(6y - 1)

Factorise x² + 2x - 8
We need to split the 2x into two numbers which multiply to give -8. This has to be 4 and -2.
x² + 4x - 2x - 8
x(x + 4) - 2x - 8
x(x + 4)- 2(x + 4)
(x + 4)(x - 2)

Once you work out what is going on, this method makes factorising any expression easy. It is worth studying these examples further if you do not understand what is happening. Unfortunately, the only other method of factorising is by trial and error.

The Difference of Two Squares

If you are asked to factorise an expression which is one square number minus another, you can factorise it immediately. This is because a² - b² = (a + b)(a - b) .

Factorise 25 - x²
= (5 + x)(5 - x) [imagine that a = 5 and b = x]

## How to Factor a Quadratic Equation?

Factoring a quadratic equation can be defined as the process of breaking the equation into the product of its factors. In other words, we can also say that factorization is the reverse of multiplying out.

To solve the quadratic equation ax 2 + bx + c = 0 by factorization, the following steps are used:

• Expand the expression and clear all fractions if necessary.
• Move all terms to the left-hand side of the equal to sign.
• Factorize the equation by breaking down the middle term.
• Equate each factor to zero and solve the linear equations

Expand the equation and move all the terms to the left of the equal sign.

Equate each factor equal to zero and solve

Therefore, the solutions are x = 2, 1/2.

Solve the following quadratic equation (2x – 3) 2 = 25

Expand the equation (2x – 3) 2 = 25 to get

Divide each term by 4 to get

The are many methods of factorizing quadratic equations. In this article, our emphasis will be based on how to factor quadratic equations, in which the coefficient of x 2 is either 1 or greater than 1.

Therefore, we will use the trial and error method to get the right factors for the given quadratic equation.

## MathHelp.com Looking at two generic binomials (using the variable x and the generic numbers p and q ), we can multiply the binomials like this:

In the above, (p + q) = b and pq = c from " x 2 + bx + c ". This multiplication and simplification explains why, to factor a quadratic, we'll need to start by finding the two numbers (being the p and the q above) that add up to equal b , where those numbers also multiply to equal c . It's required by the logic of factoring (and factoring the quadratic is the "undo" of the original binomial multiplication).

(By the way, I call this topic "factoring quadratics", where your textbook may refer to this topic as "factoring trinomials". But a "trinomial" is any three-term polynomial, which may not be a quadratic (that is, a degree-two) polynomial. And not all quadratics have three terms. So the book's section or chapter title is, at best, a bit off-target. Don't worry about the difference, though the book's title means the same thing as what this lesson explains.)

Here's what the "simple" quadratic-factorization process looks like, in practice:

#### Factor x 2 + 5x + 6

This quadratic has a leading coefficien of 1 , so this is the simple case of factoring. To get started, I need to find factors of c = +6 that add up to b = +5 . I have two options, because 6 factors as the product of 2 and 3 , or as the product of 1 and 6 .

Now, because I'm multiplying to a positive six, then my factors have to have the same sign they must both be positive or else they must both be negative, because that's how negatives work. Because I'm adding to a positive five, then both factors must be positive.

I'll check the sums of the pairs of potential factors, to see which works:

Since I need my factors to sum to plus-five, then I'll be using the factors +2 and +3 .

From back when I learned how to multiply polynomials, I know that they got this quadratic by multiplying two binomials. Because the leading coefficient is just 1 , I know that the leading coefficient of each of those binomials must also have been just 1 . This means that the product started out looking something like this:

At the ends of each parenthetical go the numbers that multiply to +6 and add to +5 . This means that I can finish my factoring by plugging those numbers into my parentheticals, in either order:

Why "in either order"? Because the front ends of the two parentheticals were the same, so I'd have ended up with the same factorization, either way. Remember: order doesn't matter in multiplication.

This is how all of the "easy" quadratics will work: we find factors of the constant term that add up to the middle term, and then we use these factors to fill in our parentheses.

By the way, we can always check our work by multiplying our factors back together, and check that we have gotten back to the original answer. To check the above factorization, the multiplication goes like this:

Your text or teacher may refer to factoring "by grouping", which is covered in the lesson on simple factoring. In the "easy" case of factoring, using the "grouping" method just gives you some extra work. For instance, in the above problem, in addition to finding the factors of +6 that add to +5 , you would have had to do these additional steps:

You would get the same answer as I did, but (I think) it's easier to just go straight to filling in the parentheses.

#### Factor x 2 + 7x + 6

The leading term is just 1 , so this is the simple case of factoring. The constant term is 6 , which can be written as the product of 2 and 3 or of 1 and 6 , just as in the previous exercise. But the coefficient on the middle term is different this time. Instead of a +5 , I've got a +7 .

The sign on the constant term is the same as before (namely, a "plus"), so I'll still need to "plus" factors. But the sum (that is, the middle-term's coefficient) is different it's now a 7 . Whereas 2 + 3 = 5 worked for the previous quadratic, +2 and +3 are not the numbers I need in this case. On the other hand, 1 + 6 = +7 , so I'll use +1 and +6 for my factorization. And I'll also go straight from having drawn this conclusion to writing down my final answer:

Again, remember that the order doesn't matter in multiplication, so the above answer could equally-correctly be written as " (x + 6)​(x + 1) ".

#### Factor x 2 &ndash 5x + 6

The constant term (created by multiplication) is +6 , so my factors will be either both "plus" or else both "minus". But the middle coefficient this time is "minus". Since I'm adding to a "minus", (namely, to &ndash5 ), then both factors must be "minus".

When the middle term's coefficient had been a "plus" five, I'd used factors +2 and +3 . Now that my middle term's coefficient is "minus", I will use &ndash2 and &ndash3 :

Note that we can use clues from the signs to determine which pairs of factors to use, as I demonstrated in the previous exercises. The rules, formally, look like this:

• If c is "plus", then the factors will be either both "plus" or else both "minus".
• If b is "plus", then the factors are both "plus".
• If b is "minus", then the factors are both "minus".
• If c is "minus", then the factors will be of alternating signs that is, one will be "plus" and one will be "minus".
• If b is "plus", then the larger of the two factors is "plus".
• If b is "minus", then the larger of the two factors is "minus".

#### Factor x 2 &ndash 7x + 6

The leading coeficient is 1 , so this is a simple-factoring quadratic. I am multiplying to a "plus" six, so the factors will be either both "plus" or else both "minus". Looking at the middle term, I see that I am adding to a "minus" seven, so my factors will both be "minus".

The (negative) factors of +6 that add up to 7 are &ndash1 and &ndash6 , so I will use &ndash1 and &ndash6 for my factorization:

So far, c (the constant term) has always been a "plus". What if c is a "minus"?

#### Factor x 2 + x &ndash 6

Since I am multiplying to a "minus" six, I need factors that have opposite signs that is, one factor will be "plus" and the other will be "minus". The coefficient on the middle term is a "plus" 1 , so I know that the larger of the two factors (larger, that is, in terms of its absolute value) will get the "plus" sign. Since these opposite-signed numbers will be added together to get +1 , then I need the two factors to be one unit apart.

The factor pairs for 6 are 1 and 6 , and 2 and 3 . The values in the second pair are one unit apart, so I know that I will be using 2 and 3 .

Because I need to get a "plus" answer for the sum of the two factors, I'll need the larger of my two numbers to get the "plus" sign namely, the 3 will be getting the "plus" sign (so the 2 gets the "minus" sign). Then my factorization is:

#### Factor x 2 &ndash x &ndash 6

This looks just like the previous quadratic, except that now the middle term is "minus". The constant term is still "minus", so I still want factors with opposite signs. And the middle term's coefficient (other than its sign) is still 1 , so I still want factors that are one unit apart. But this time the larger factor will get the "minus" sign.

All the other considerations remain the same. I still want factors of 6 that are one unit apart, so I'll still be using 2 and 3 . But this time, the sign on the 3 will be a "minus":

## 6: Factoring - Mathematics

16. Factor the following polynomial.

Show All Steps Hide All Steps

Don’t let the fact that this polynomial is not quadratic worry you. Just because it’s not a quadratic polynomial doesn’t mean that we can’t factor it.

For this polynomial we can see that (> ight)^2> = ) and so it looks like we can factor this into the form,

At this point all we need to do is proceed as we did with the quadratics we were factoring above.

After writing down the factors of -4 we can see that we need to have the following factoring.

[ + 3 - 4 = left( <+ 4> ight)left( <- 1> ight)] Show Step 3

Now, we need to be careful here. Sometimes these will have further factoring we can do. In this case we can see that the second factor is a difference of perfect cubes and we have a formula for factoring a difference of perfect cubes.

Therefore, the factoring of this polynomial is,

[ + 3 - 4 = left( <+ 4> ight)left( <- 1> ight) = equire box[2pt,border:1px solid black] <+ 4> ight)left( ight)left( <+ x + 1> ight)>>]

## MathHelp.com If I don't first take out the common factor of " 2 ", I'll find the factors of 2×(&ndash16) = &ndash32 that add to &ndash4 . In other words, I'll need factors of opposite sign (so I'll be subtracting them) which are four units apart.

The factor pairs of 32 , and their differences, are:

Okay so 8 and 4 are four units apart. Since I'm adding to a "minus", I'll slap the "minus" sign on the larger of the two numbers. I'll be using &ndash8 and +4 to split the middle term's coefficient of &ndash4 . My "box" looks like this:

My factorization is (2x &ndash 8)(2x + 4) , which I can check by multiplying this back together. But right at the start of multiplying this back out, I see that I'm getting a leading term of 4x 2 , which is not what I'd started with. So clearly this is wrong!

By not taking that common factor out first, I have managed to create extra factors in "box" in particular, by not pulling the 2 out front first, I've created an extraneous factor of 2 .

To do things properly, I start by pulling out that common factor of 2 . This gives me:

Now I need to factor the remaining quadratic:

Oh, hey! After pulling out the common factor, this turned into one of the simple-case factorizations! All I need are factors of &ndash8 that sum to &ndash2 . In other words, I need factors of 8 that are two units apart, where the larger of the factors (other than sign) gets the "minus" sign. That's easy I'll use &ndash4 and +2 . The quadratic factors as:

Then my answer (remembering to include the 2 that I'd factored out at the start) is:

The one special case that often causes students some trouble is when the leading coeffiecient is a "minus" value. A good first step is to factor that value out of the entire quadratic (or, at least factor the "minus" out of the whole thing).

#### Factor &ndash6x 2 &ndash x + 2

There are no (non-trivial) common factors, so there's nothing "interesting" (like a 2 ) that I can pull out of all three terms. The leading coefficient is not 1 , so I will need to use "box". Because the leading coefficient is also "minus", my first step will be to pull a &ndash1 from all three terms. This gives me:

(I need to remember that every sign changes when I multiply or divide through by a "minus". I mustn't fall into the trap of taking the &ndash1 out of only the first term I must take it out of all three!)

I can see that I'll need factors of ac = (6)(&ndash2) = &ndash12 (so one "plus" and one "minus") that add to the middle term's coefficient of 1 (so the factors will be one unit apart, with the larger one getting the "plus" sign). This one is simple enough that I don't need to list factor pairs I already know that 3 and 4 are one unit apart, so I'll split the 1 by using +4 and &ndash3 .

Plugging the contents of the parentheses into "box" gives me:

Remembering the &ndash1 that I factored out front in my first step, my answer then is:

By putting these two techniques together (that is, by factoring out anything common to all three terms, and by taking out a leading "minus" sign), we can now handle factoring messier quadratics, such as:

#### Factor &ndash6x 2 + 15x + 36

The leading coefficient is not 1 , so I'll be needing to use "box" to factor, as things stand now. However, I can see that there is a factor of 3 that's common to all three terms, and also that the leading coefficient is a "minus" this tells me that a good first step will be to pull a &ndash3 out front, applying "box" to whatever is left. So my first step gives me:

Now I need to factor the remaining quadratic. I need to find factors of ac = (2)(&ndash12) = &ndash24 (so one will be "plus" and the other will be "minus") that add to &ndash5 (so they'll be five units apart, and the larger factor will get the "minus" sign). Assuming I didn't notice the required factor pair right away, here are the pairs, and their differences:

There is another factor pair, but I can stop at this point, because I've got the pair I need namely, 3 and 8 . Because I need them to have opposite signs and to add to a "minus", I will put the "minus" sign on the 8 . Then "box" looks like this:

When I write down my answer, I need to remember to include the &ndash3 that I factored out front in my first step:

A disguised version of this factoring-out-the-"minus" case is when they give us a backwards quadratic where the squared term is subtracted, like this:

To do the factorization, the first step would be to reverse the quadratic to put it back in the "normal" order

Then we'd factor in the usual way:

We can do this "swapping the terms around" in the above case because order doesn't matter in addition. In subtraction, however, order does matter when flipping around a quadratic with "minus" signs, we need to be careful with those signs. For instance:

#### Factor 6 + x &ndash x 2

The "leading" term isn't up front, so I'll want to start by flipping things around. However, I'll need to take care with the signs, making sure that I carry them along with their (following) terms. This gives me:

This gives me a quadratic with a "minus" leading coefficient, so I'll factor a &ndash1 out of all three terms:

This leaves me with a simple quadratic to factor. The factors of &ndash6 (so one is "plus" and the other is "minus") add up to &ndash1 (so the factors are one unit apart, and the larger one gets the "minus" sign) clearly, I should use +2 and &ndash3 . Including the &ndash1 that I pulled out front, this gives me:

Usually, they'll give us quadratics for which the product ab doesn't have too terribly many factor pairs. However, sometimes ab is large enough that it's painful to find the necessary factor pair. In such a case, we can always create a list (using our calculator to divide progressively larger values into ab to find those pairs), quitting when we find the pair we need. That process looks like this:

#### Factor 20x 2 &ndash 17x &ndash 63

For this quadratic, ac = (20)(&ndash63) = &ndash1260 . Off the top of my head, I have no frikkin' idea what factors I'll need to use. All I know so far is that those factors will have opposite signs, and that they'll be seventeen units apart, with the larger of the two getting the "minus" sign. So I'll start listing pairs of factors of 1,260 , and see if I can find a pair that works.

My calculator tells me that 8 isn't a factor of 1,260 , but 9 is (and I can plainly see that 10 is), so:

Skipping the remaining numbers that my calculator tells me aren't factors, I come up with:

Finally! Yes, there are two more factor pairs for 1,260 , but I've finally found a pair that works, so I can stop here. Since I'm adding to a "minus", I'll be putting the "minus" sign on the larger of the two factors, and my "box" looks like this:

You should expect an exercise as long as this on the next test. When you get a value for ac that's ridiculously large, don't waste a lot of time trying to "eyeball" the solution. When you have numbers this big, it can actually be faster to write down the list of factor pairs. In such a case, write your pairs in your hand-in work, as this will be important evidence that you found the factorization, rather than your calculator or your phone (or "your neighbor's paper").

There is one other type of quadratic that looks kind of different, but the factoring itself works in exactly the same way it's the case of a quadratic with two squared variables, one at each end of the quadratic:

#### Factor 6x 2 + xy &ndash 12y 2

This may look bad, what with the y 2 at the end, but it factors just like all the previous quadratics. I recall, from simple factoring, that I know that the factors of a simple-case quadratic had to be in the following form:

(x + something )(x + something else )

For this slightly-different quadratic, they must have multiplied factors in this slightly-different form:

So I'll have y 's at the ends of each of my parentheticals. But other than this, the process will work as usual.

First, I need to find factors of ac = (6)(&ndash12) = &ndash72 (so one "plus" and one "minus") that add to +1 (so they'll be one unit apart, with the larger one getting the "plus" sign). This factorization is easy I'll use +9 and &ndash8 to split the middle term. So "box" looks like this:

You can use the Mathway widget below to practice factoring quadratics (or, as the widget calls them, "trinomials"). Try the entered exercise, or type in your own exercise. (Or skip the widget and continue on the next page.) Then click the button to compare your answer to Mathway's.

They also have an option for checking if a polynomial (such as a quadratic) is "prime" enter the quadratic, click the button, and select "Determine if Prime".)

(Click "Tap to view steps" to be taken directly to the Mathway site for a paid upgrade.)

## Common Factor Calculator

Please provide integers separated by a comma "," and click the "Calculate" button to find their common factors.

### What is a factor?

A factor is a term in multiplication. For example, in:

3 and 4 are the factors. It is possible for a number to have multiple factors. Using 12 as an example, in addition to 3 and 4 being factors:

It can be seen that 1, 2, 3, 4, 6, and 12 are all factors of the number 12. This is the most basic form of a factor, but algebraic expressions can also be factored, though that is not the intent of this calculator.

### What is a common factor?

A common factor is a factor that is shared between two different numbers. It can also be referred to as a common divisor. As an example:

The factors of 16 include: 1, 2, 4, 8, and 16.

The factors of 12 include: 1, 2, 3, 4, 6, and 12.

Thus, the common factors of 16 and 12 are: 1, 2, and 4.

Often in math problems, it can be desirable to find the greatest common factor of some given numbers. In this case, the greatest common factor is 4.

This calculator only accepts positive integers as input to calculate their common factors. While only two numbers are used in the above example, the calculator can compute the common factors of more than two numbers.

## Factor Trinomials by Unfoiling (Trial and Error)

One of the methods that we can use to factor trinomials is by trial and error or unfoiling or reverse FOIL.

In these lessons, we will learn how to factorize trinomials by the trial and error method. Many examples and worked solutions are shown.

It is also possible to factorize quadratic trinomials without trial and error. This is shown in the last video on this page.

Factor the following trinomial.

Step 1: The first term is x 2 , which is the product of x and x. Therefore, the first term in each bracket must be x, i.e.

Step 2 : The last term is 6. The possible factors are ±1 and ±6 or ±2 and ±3. So, we have the following choices.

The only pair of factors which gives -5x as the middle term is (x - 3)(x - 2)

Step 3: The answer is then

How to factor trinomials by reverse FOIL (or trial and error method) and Grouping?
Factor ax 2 + bx + c where a &ne 1.
Method 1: Trail and Error (Reverse FOIL)
Step 1: Place the factors of ax 2 in the first positions of the 2 sets of parentheses that represent the factors.
Step 2: Place 2 possible factors of c into the second positions of the parentheses.
Step 3: Find the inner and outer products of the 2 sets of parentheses.
Step 4: Keep trying different factors until the inner and outer products add to bx.
Examples:
Factor
1. 6x 2 - 19x + 15
2. 15x 2 + 17x - 42 Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. ## Rs Aggarwal for Class 6 Math Chapter 2 - Factors And Multiples

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#### Question 1:

Define: (i) factor (ii) multiple. Give five examples of each.

Factor: A factor of a number is an exact divisor of that number.
Multiple: A multiple of a number is a number obtained by multiplying it by a natural number.

Example 1: We know that 15 = 1 × 15 and 15 = 3 × 5

&there4 1, 3, 5 and 15 are the factors of 15
In other words, we can say that 15 is a multiple of 1, 3, 5 and 15.

Example 2: We know that 8 = 8 × 1, 8 = 2 × 4 and 8 = 4 × 2

&there4 1, 2, 4 and 8 are the factors of 8.
In other words, we can say that 8 is a multiple of 1, 2, 4 and 8.

Example 3: We know that 30 = 30 × 1, 30 = 5 × 6 and 30 = 6 × 5

&there4 1, 5, 6 and 30 are factors of 30.
In other words, we can say that 30 is a multiple of 1, 5, 6 and 30.

Example 4: We know that 20 = 20 × 1, 20 = 4 × 5 and 20 = 5 × 4

&there4 1, 4, 5 and 20 are factors of 20.
In other words, we can say that 20 is a multiple of 1, 4, 5 and 20.

Example 5: We know that 10 = 10 × 1, 10 = 2 × 5 and 10 = 5 × 2

&there4 1, 2, 5 and 10 are factors of 10.
In other words, we can say that 10 is a multiple of 1, 2, 5 and 10.

#### Question 2:

Write down all the factors of
(i) 20
(ii) 36
(iii) 60
(iv) 75

(i) 20
20 = 1 × 20 20 = 10 × 2 and 20 = 4 × 5
The factors of 20 are 1, 2, 4, 5, 10 and 20.

(ii) 36
36 = 1 × 36 36 = 2 × 18 36 = 3 × 12 and 36 = 4 × 9
The factors of 36 are 1, 2, 3, 4, 6, 9, 12 and 36.

(iii) 60
60 = 1 × 60 60 = 2 × 30 60 = 3 × 20 60 = 4 × 15 and 60 = 5 × 12
The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15 and 60.

(iv) 75
75 = 1 × 75 75 = 3 × 25 and 75 = 5 × 15
The factors of 75 are 1, 3, 5, 15, 25 and 75.

#### Question 3:

Write the first five multiples of each of the following numbers:
(i) 17
(ii) 23
(iii) 65
(iv) 70

(i) 17
17 × 1 = 17 17 × 2 = 34 17 × 3 = 51 17 × 4 = 68 and 17 × 5 = 85
&there4 The first five multiples of 17 are 17, 34, 51, 68 and 85.

(ii) 23
23 × 1=23 23 × 2 = 46 23 × 3 = 69 23 × 4 = 92 and 23 × 5 = 115
&there4 The first five multiples of 23 are 23, 46, 69, 92 and 115.

(iii) 65
65 × 1 = 65 65 × 2 = 130 65 × 3 = 195 65 × 4 = 260 and 65 × 5 = 325
&there4 The first five multiples of 65 are 65, 130, 195, 260 and 325.

(iv) 70
70 × 1=70 70 × 2 = 140 70 × 3 = 210 70 × 4 = 280 and 70 × 5 = 350
&there4 The first five multiples of 70 are 70, 140, 210, 280 and 350.

#### Question 4:

Which of the following numbers are even and which are odd?
(i) 32
(ii) 37
(iii) 50
(iv) 58
(v) 69
(vi) 144
(vii) 321
(viii) 253

(i) 32
Since 32 is a multiple of 2, it is an even number.
(ii) 37
Since 37 is not a multiple of 2, it is an odd number.
(iii) 50
Since 50 is a multiple of 2, it is an even number.
(iv) 58
Since 58 is a multiple of 2, it is an even number.
(v) 69
Since 69 is not a multiple of 2, it is an odd number.
(vi) 144
Since 144 is a multiple of 2, it is an even number.
(vii) 321
Since 321 is not a multiple of 2, it is an odd number.
(viii) 253
Since 253 is not a multiple of 2, it is an odd number.

#### Question 5:

What are prime numbers? Give ten examples.

Prime number: A number is called a prime number if it has only two factors, namely 1 and itself .

Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29 are prime numbers.

#### Question 6:

Write all the prime numbers between
(i) 10 and 40
(ii) 80 and 100
(iii) 40 and 80
(iv) 30 and 40

(i) All prime numbers between 10 and 40 are 11, 13, 17, 19, 23, 29, 31 and 37.
(ii) All prime numbers between 80 and 100 are 83, 89 and 97.
(iii) All prime numbers between 40 and 80 are 41, 43, 47, 53, 59, 61, 67, 71, 73 and 79.
(iv) All prime numbers between 30 and 40 are 31 and 37.

#### Question 7:

(i) Write the smallest prime number.
(ii) List all even prime numbers.
(iii) Write the smallest odd prime number.

(i) The smallest prime number is 2.
(ii) There is only one even prime number, i.e., 2.
(iii) The smallest odd prime number is 3.

#### Question 8:

Find which of the following numbers are primes:
(i) 87
(ii) 89
(iii) 63
(iv) 91

(i) 87
The divisors of 87 are 1, 3, 29 and 87 i.e. 87 has more than 2 factors. Therefore 87 is not a prime number.

(ii) 89
The divisors of 89 are 1 and 89. Therefore 89 is a prime number.

(iii) 63
The divisors of 63 are 1, 3, 7, 9, 21 and 63 i.e. 63 has more than 2 factors. Therefore 63 is not a prime number.

(iv) 91
The divisors of 91 are 1, 7, 13 and 91 i.e. 91 has more than 2 factors. Therefore 91 is not a prime number.

#### Question 9:

Make a list of seven consecutive numbers, none of which is prime.

90, 91, 92, 93, 94, 95 and 96 are seven consecutive numbers and none of them is a prime.

#### Question 10:

(i) Is there any counting number having no factor at all?
(ii) Find all the numbers having exactly one factor.
(iii) Find numbers between 1 and 100 having exactly three factors.

(i) No, there are no counting numbers with no factors at all because every number has at least two factors, i.e., 1 and itself.
(ii) There is only one number that has exactly one factor, i.e, 1.
(iii) The numbers between 1 and 100 that have exactly three factors are 4, 9, 25 and 49.

#### Question 11:

What are composite numbers? Can a composite number be odd? If yes, write the smallest odd composite number.

The numbers that have more than two factors are known as composite numbers.
Yes, a composite number can be odd.
The smallest odd composite number is 9.

#### Question 12:

What are twin primes? Write all the pairs of twin primes between 50 and 100.

Two consecutive odd prime numbers are called twin primes.
The pairs of twin primes between 50 to 100 are (59, 61) and (71, 73).

#### Question 13:

What are co-primes? Give examples of five pairs of co-primes. Are co-primes always primes? If no, illustrate your answer by an example.

If two numbers do not have a common factor other than 1, they are said to be co-primes.

Five pairs of co primes: (i) 2 and 3 (ii) 3 and 4 (iii) 4 and 5 (iv) 4 and 9 (v) 8 and 15

No, co&ndashprimes are not always primes.

For example, 3 and 4 are co-prime numbers, where 3 is a prime number and 4 is not a prime number.

#### Question 14:

Express each of the following numbers as the sum of two odd primes:
(i) 36
(ii) 42
(iii) 84
(iv) 98

(i) 36
36 as the sum of two odd prime numbers is (36 = 31 + 5).
(ii) 42
42 as the sum of two odd prime numbers is (42 = 31 + 11).
(iii) 84
84 as the sum of two odd prime numbers is (84 = 41 + 43).
(iv) 98
98 as the sum of two odd prime numbers is (98 = 31 + 67).

#### Question 15:

Express each of the following odd numbers as the sum of three odd prime numbers:
(i) 31
(ii) 35
(iii) 49
(iv) 63

(i) 31
31 can be expressed as the sum of three odd prime numbers as (31 = 5 + 7 + 19).
(ii) ) 35
35 can be expressed as the sum of three odd prime numbers as (35 = 17 + 13 + 5).
(iii) 49
49 can be expressed as the sum of three odd prime numbers as (49 = 13 + 17 + 19).
(iv) 63
63 can be expressed as the sum of three odd prime numbers as (63 = 29 + 31 + 3).

#### Question 16:

Express each of the following numbers as the sum of twin primes:
(i) 36
(ii) 84
(iii) 120
(iv) 144

(i) 36
36 can be expressed as the sum of twin primes as (36 = 17 + 19).
(ii) 84
84 can be expressed as the sum of twin primes as (84 = 41 + 43).
(iii) 120
120 can be expressed as the sum of twin primes as (120 = 59 + 61).
(iv) 144
144 can be expressed as the sum of twin primes as (144 = 71 + 73).

#### Question 17:

Which of the following statements are true?
(i) 1 is the smallest prime number.
(ii) If a number is prime, it must be odd.
(iii) The sum of two prime numbers is always a prime number.
(iv) If two numbers are co-primes, at least one of them must be a prime number.

(i) False. 2 is the smallest prime number.
(ii) False. 2 is an even prime number.
(iii) False. 3 and 7 are two prime numbers and their sum is 10, which is even.
(iv) False. 4 and 9 are co-primes but neither of them is a prime number.

#### Question 1:

Test the divisibility of the following numbers by 2:
(i) 2650
(ii) 69435
(iii) 59628
(iv) 789403
(v) 357986
(vi) 367314

A number is divisible by 2 if its ones digit is 0, 2, 4, 6 or 8.
(i) Since the digit in the ones place in 26250 is 0, it is divisible by 2
(ii) Since the digit in the ones place in 69435 is not 0, 2, 4, 6 or 8, it is not divisible by 2.
(iii) Since the digit in the ones place in 59628 is 8, it is divisible by 2.
(iv) Since the digit in the ones place in 789403 is not 0, 2, 4, 6, or 8, it is not divisible by 2.
(v) Since the digit in the ones place in 357986 is 6, it is divisible by 2.
(vi) Since the digit in the ones place in 367314 is 4, it is divisible by 2.

#### Question 2:

Test the divisibility of the following numbers by 3:
(i) 733
(ii) 10038
(iii) 20701
(iv) 524781
(v) 79124
(vi) 872645

A number is divisible by 3 if the sum of its digits is divisible by 3.
(i) 733 is not divisible by 3 because the sum of its digits, 7 + 3 + 3, is 13, which is not divisible by 3.
(ii) 10038 is divisible by 3 because the sum of its digits, 1 + 0 + 0 + 3 + 8, is 12, which is divisible by 3.
(iii) 20701 is not divisible by 3 because the sum of its digits, 2 + 0 + 7 + 0 + 1, is 10, which is not divisible by 3.
(iv) 524781 is divisible by 3 because the sum of its digits, 5 + 2 + 4 + 7 + 8 + 1, is 27, which is divisible by 3.
(v) 79124 is not divisible by 3 because the sum of its digits, 7 + 9 + 1 + 2 + 4, is 23, which is not divisible by 3.
(vi) 872645 is not divisible by 3 because the sum of its digits, 8 + 7 + 2 + 6 + 4 + 5, is 32, which is not divisible by 3.

#### Question 3:

Test the divisibility of the following numbers by 4:
(i) 618
(ii) 2314
(iii) 63712
(iv) 35056
(v) 946126
(vi) 810524

A number is divisible by 4 if the number formed by the digits in its tens and units place is divisible by 4.

(i) 618 is not divisible by 4 because the number formed by its tens and ones digits is 18, which is not divisible by 4.
(ii) 2314 is not divisible by 4 because the number formed by its tens and ones digits is 14, which is not divisible by 4.
(iii) 63712 is divisible by 4 because the number formed by its tens and ones digits is 12, which is divisible by 4.
(iv) 35056 is divisible by 4 because the number formed by its tens and ones digits is 56, which is divisible by 4.
(v) 946126 is not divisible by 4 because the number formed by its tens and ones digits is 26, which is not divisible by 4.
(vi) 810524 is divisible by 4 because the number formed by its tens and ones digits is 24, which is divisible by 4.

#### Question 4:

Test the divisibility of the following numbers by 5:
(i) 4965
(ii) 23590
(iii) 35208
(iv) 723405
(v) 124684
(vi) 438750

A number is divisible by 5 if its ones digit is either 0 or 5.

(i) 4965 is divisible by 5, because the digit at its ones place is 5.

(ii) 23590 is divisible by 5, because the digit at its ones place is 0.

(iii) 35208 is not divisible by 5, because the digit at its ones place is 8.

(iv) 723405 is divisible by 5, because the digit at its ones place is 5.

(v) 124684 is not divisible by 5, because the digit at its ones place is 4.

(vi) 438750 is divisible by 5, because the digit at its ones place is 0.

#### Question 5:

Test the divisibility of the following numbers by 6:
(i) 2070
(ii) 46523
(iii) 71232
(iv) 934706
(v) 251780
(vi) 872536

A number is divisible by 6 if it is divisible by both 2 and 3.

i) Since 2070 is divisible by 2 and 3, it is divisible by 6.
Checking the divisibility by 2: Since the number 2070 has 0 in its units place, it is divisible by 2.
Checking the divisibility by 3: The sum of the digits of 2070, 2 + 0 + 7 + 0, is 9, which is divisible by 3. So, it is divisible by 3.

(ii) Since 46523 is not divisible by 2, it is not divisible by 6.

Checking the divisibility by 2: Since the number 46523 has 3 in its units place, it is not divisible by 2.

(iii) Since 71232 is divisible by both 2 and 3, it is divisible by 6.

Checking the divisibility by 2: Since the number has 2 in its units place, it is divisible by 2.
Checking the divisibility by 3: The sum of the digits of the number, 7 + 1 + 2 + 3 + 2, is 15, which is divisible by 3. So, the number is divisible by 3.

(iv) Since 934706 is not divisible by 3, it is not divisible by 6.
Checking the divisibility by 3: Since the sum of the digits of the number, 9 + 3 + 4 + 7 + 0 + 6, is 29, which is not divisible by 3. So, the numbe r is not divisible by 3.

(v) Since 251780 is not divisible by 3, it is not divisible by 6.
Checking the divisibility by 3: The sum of the digits of the number, 2 + 5 + 1 + 7 + 8 + 0, is 23, which is not divisible by 3. So, the numbe r is not divisible by 3.

(vi) Since 872536 is not divisible by 3 , it is not divisible by 6.
Checking the divisibility by 3: The sum of the digits of the number, 8 + 7 + 2 + 5 + 3 + 6, is 31, which is not divisible by 3. So, the numbe r is not divisible by 3.

#### Question 6:

Test the divisibility of the following numbers by 7:
(i) 826
(ii) 117
(iii) 2345
(iv) 6021
(v) 14126
(vi) 25368

To determine if a number is divisible by 7, double the last digit of the number and subtract it from the number formed by the remaining digits. If their difference is a multiple of 7, the number is divisible by 7.

(i) 826 is divisible by 7.
We have 82 &minus 2 × 6 = 70, which is a multiple of 7.

(ii) 117 is not divisible by 7.
We have 11 &minus 2 × 7 = &minus3, which is not a multiple of 7.

(iii) 2345 is divisible by 7.
We have 234 &minus 2 × 5 = 224, which is a multiple of 7.

(iv) 6021 is divisible by 7.
We have 602 &minus 2 × 1 = 600, which is not a multiple of 7.

(v) 14126 is divisible by 7.
We have 1412 &minus 2 × 6 = 1400, which is a multiple of 7.

(vi) 25368 is divisible by 7.
We have 2536 &minus 2 × 8 = 2520, which is a multiple of 7.

#### Question 7:

Test the divisibility of the following numbers by 8:
(i) 9364
(ii) 2138
(iii) 36792
(iv) 901674
(v) 136976
(vi) 1790184

A number is divisible by 8 if the number formed by the last three digits (digits in the hundreds, tens and units places) is divisible by 8.

(i) 9364 is not divisible by 8.
It is because the number formed by its hundreds, tens and ones digits, i.e., 364, is not divisible by 8.

(ii) 2138 is not divisible by 8.
It is because the number formed by its hundreds, tens and ones digits, i.e., 138, is not divisible by 8.

(iii) 36792 is divisible by 8.
It is because the number formed by its hundreds, tens and ones digits, i.e., 792, is divisible by 8.

(iv) 901674 is not divisible by 8.
It is because the number formed by its hundreds, tens and ones digits, i.e., 674, is not divisible by 8.

(v) 136976 is divisible by 8.
It is because the number formed by its hundreds, tens and ones digits, i.e., 976, is divisible by 8.

(vi) 1790184 is divisible by 8.
It is because the number formed by its hundreds, tens and ones digits, i.e., 184, is divisible by 8.

#### Question 8:

Test the divisibility of the following numbers by 9:
(i) 2358
(ii) 3333
(iii) 98712
(iv) 257106
(v) 647514
(vi) 326999

A number is divisible by 9 if the sum of its digits is divisible by 9.

(i) 2358 is divisible by 9, because the sum of its digits, 2 + 3 + 5 + 8, is 18, which is divisible by 9.

(ii) 3333 is not divisible by 9, because the sum of its digits, 3 + 3 + 3 + 3, is 12, which is not divisible by 9.

(iii) 98712 is divisible by 9, because the sum of its digits, 9 + 8 + 7 + 1 + 2, is 27, which is divisible by 9.

(iv) 257106 is not divisible by 9, because the sum of its digits, 2 + 5 + 1 0 + 6, is 21, which is not divisible by 9.

(v) 647514 is divisible by 9, because the sum of its digits, 6 + 4 + 7 + 5 + 1 + 4, is 27, which is divisible by 9.

(vi) 326999 is not divisible by 9, because the sum of its digits, 3 + 2 + 6 + 9 + 9 + 9, is 38, which is not divisible by 9.

#### Question 9:

Test the divisibility of the following numbers by 10:
(i) 5790
(ii) 63215
(iii) 55555

A number is divisible by 10 if its ones digit is 0.

(i) 5790 is divisible by 10, because its ones digit is 0.
(ii) 63215 is not divisible by 10, because its ones digit is 5, not 0.
(iii) 55555 is not divisible by 10, because its ones digit is 5, not 0.

#### Question 10:

Test the divisibility of the following numbers by 11:
(i) 4334
(ii) 83721
(iii) 66311
(iv) 137269
(v) 901351
(vi) 8790322

A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.

(i) 4334 is divisible by 11.

Sum of the digits at odd places = (4 + 3) = 7
Sum of the digits at even places = (3 + 4) = 7
Difference of the two sums = (7 &minus 7) = 0, which is divisible by 11 .

(ii) 83721 is divisible by 11 .
Sum of the digits at odd places = (1 + 7 + 8) = 16
Sum of the digits at even places = (2 + 3) = 5
Difference of the two sums = (16 &minus 5) = 11, which is divisible by 11.

(iii) 66311 is not divisible by 11 .

Sum of the digits at odd places = (1 + 3 + 6) = 10
Sum of the digits at even places = (1 + 6) = 7
Difference of the two sums = (10 &minus 7) = 3, which is not divisible by 11.

(iv) 137269 is divisible by 11 .

Sum of the digits at odd places = (9 + 2 + 3) = 14
Sum of the digits at even places = (6 + 7 + 1) = 14
Difference of the two sums = (14 &minus 14) = 0, which is a divisible by 11.

(v) 901351 is divisible by 11 .

Sum of the digits at odd places = (0 + 3 + 1) = 4
Sum of the digits at even places = (9 + 1 + 5) = 15
Difference of the two sums = (4 &minus 15) = &minus11, which is divisible by 11.

(vi) 8790322 is not divisible by 11 .

Sum of the digits at odd places = (2 + 3 + 9 + 8) = 22
Sum of the digits at even places = (2 + 0 + 7) = 9
Difference of the two sums = (22 &minus 9) = 13, which is not divisible by 11.

#### Question 11:

In each of the following numbers, replace * by the smallest number to make it divisible by 3:
(i) 27*4
(ii) 53*46
(iii) 8*711
(iv) 62*35
(v) 234*17
(vi) 6*1054

(i) 2724
Here, 2 + 7 + * + 4 = 13 + * should be a multiple of 3.
To be divisible by 3, the least value of * should be 2, i.e., 13 + 2 = 15, which is a multiple of 3.
&there4 * = 2

(ii) 53046
Here, 5 + 3 + * + 4 + 6 = 18 + * should be a multiple of 3.
As 18 is divisible by 3, the least value of * should be 0, i.e., 18 + 0 = 18.
&there4 * = 0

(iii) 81711
Here, 8+ * + 7 + 1 + 1 = 17 + * should be a multiple of 3.
To be divisible by 3, the least value of * should be 1, i.e., 17 + 1 = 18 , which is a multiple of 3.
&there4 * = 1

(iv) 62235
Here, 6 + 2 + * + 3 + 5 = 16 + * should be a multiple of 3.
To be divisible by 3, the least value of * should be 2, i.e., 16 + 2 = 18, which is a multiple of 3.
&there4 * = 2

(v) 234117
Here, 2+ 3 +4 + * + 1 + 7 = 17 + * should be a multiple of 3.
To be divisible by 3, the least value of * should be 1, i.e., 17 + 1 = 18, which is a multiple of 3.
&there4 * =1

(vi) 621054
Here, 6 + * +1 + 0 + 5 + 4 = 16 + * should be a multiple of 3.
To be divisible by 3, the least value of * should be 2, i.e., 16 + 2 = 18, which is a multiple of 3.
&there4 * =2

#### Question 12:

In each of the following numbers, replace * by the smallest number to make it divisible by 9:
(i) 65*5
(ii) 2*135
(iii) 6702*
(iv) 91*67
(v) 6678*1
(vi) 835*86

(i) 6525
Here, 6 + 5+ * + 5 = 16 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 2, i.e., 16 + 2 = 18, which is a multiple of 9.
&there4 * =2

(ii) 27135
Here, 2 + * + 1 + 3 + 5 = 11 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 7, i.e., 11 + 7 = 18, which is a multiple of 9.
&there4 * = 7

(iii) 67023
Here, 6 + * + 7 + 0 + 2 = 15 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 3, i.e., 15 + 3 = 18, which is a multiple of 9.
&there4 * = 3

(iv) 91467
Here, 9 + 1 * + 6 + 7 = 23 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 4, i.e., 23 + 4 = 27, which is a multiple of 9.
&there4 * = 4

(v) 667881
Here, 6 + 6 + 7 + 8 + * + 1 = 28 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 8, i.e., 28 + 8 = 36, which is a multiple of 9.
&there4 * = 8

(vi) 835686
Here, 8 + 3 + 5 + * + 8 + 6 = 30 + * should be a multiple of 9.
To be divisible of 9, the least value of * should be 6, i.e., 30 + 6 = 36, which is a multiple of 9.
&there4 * = 6

#### Question 13:

In each of the following numbers, replace * by the smallest number to make it divisible by 11:
(i) 26*5
(ii) 39*43
(iii) 86*72
(iv) 467*91
(v) 1723*4
(vi) 9*8071

(i) 26*5
Sum of the digits at odd places = 5 + 6 = 11
Sum of the digits at even places = * + 2
Difference = s um of odd terms &ndash sum of even terms
= 11 &ndash (* + 2)
= 11 &ndash * &ndash 2
= 9 &ndash *
Now, (9 &ndash *) will be divisible by 11 if * = 9.
i.e., 9 &ndash 9 = 0
0 is divisible by 11.
&there4 * = 9
Hence, the number is 2695.

(ii) 39*43
Sum of the digits at odd places = 3 + * + 3 = 6 + *
Sum of the digits at even places = 4 + 9 = 13
Difference = s um of odd terms &ndash sum of even terms
= 6 + * &ndash 13
= * &ndash 7
Now, (* &ndash 7) will be divisible by 11 if * = 7 .
i.e., 7 &ndash 7 = 0
0 is divisible by 11.
&there4 * = 7
Hence, the number is 39743.

(iii) 86*72
Sum of the digits at odd places 2 + * + 8 = 10 + *
Sum of the digits at even places 6 + 7 = 13
Difference = s um of odd terms &ndash sum of even terms
= 10 + * &ndash 13
= * &ndash 3
Now, (* &ndash 3) will be divisible by 11 if * = 3 .
i.e., 3 &ndash 3 = 0
0 is divisible by 11.
&there4 * = 3
Hence, the number is 86372.

(iv) 467*91
Sum of the digits at odd places 1 + * + 6 = 7 + *
Sum of the digits at even places 9 + 7 + 4 = 20
Difference = s um of odd terms &ndash sum of even terms
= (7 + *) &minus 20
= * &minus 13
Now, (* &minus13) will be divisible by 11 if * = 2 .
i.e., 2&minus 13 = &minus11
&minus11 is divisible by 11.
&there4 * = 2
Hence, the number is 467291.

(v) 1723*4
Sum of the digits at odd places 4+ 3+ 7= 14
Sum of the digits at even places *+2+1 = 3 + *
Difference = s um of odd terms &ndash sum of even terms
= 14 &ndash (3 + *)
= 11 &minus *
Now, (11 &minus *) will be divisible by 11 if * = 0 .
i.e., 11 &minus 0 = 11
11 is divisible by 11.
&there4 * = 0
Hence, the number is 172304.

(vi) 9*8071
Sum of the digits at odd places 1+0+* = 1 + *
Sum of the digits at even places 7 + 8 + 9 = 24
Difference = s um of odd terms &ndash sum of even terms
=1 + * &ndash 24
= * &minus 23
Now, (* &minus 23) will be divisible by 11 if * = 1.
i.e., 1 &minus 23 = &minus22
&minus22 is divisible by 11.
&there4 * = 1
Hence, the number is 918071 .

#### Question 14:

Test the divisibility of:
(i) 1000001 by 11
(ii) 19083625 by 11
(iii) 2134563 by 9
(iv) 10001001 by 3
(v) 10203574 by 4
(vi) 12030624 by 8

(i) 10000001 by 11
10000001 is divisible by 11 .
Sum of digits at odd places = (1 + 0 + 0 + 0) = 1
Sum of digits at even places = (0 + 0 + 0 + 1) = 1
Difference of the two sums = (1 &minus 1) = 0, which is divisible by 11.

(ii) 19083625 by 11
19083625 is divisible by 11 .
Sum of digits at odd places = (5 + 6 + 8 + 9) = 28
Sum of digits at even places = (2 + 3 + 0 + 1) = 6
Difference of the two sums = (28 &minus 6) = 22, which is divisible by 11.

(iii) 2134563 by 9
2134563 is not divisible by 9.
It is because the sum of its digits, 2 + 1 + 3 + 4 + 5 + 6 + 3, is 24, which is not divisible by 9.

(iv) 10001001 by 3
10001001 is divisible by 3.
It is because the sum of its digits, 1 + 0 + 0 + 0 + 1 + 0 + 0 + 1, is 3, which is divisible by 3 .

(v) 10203574 by 4
10203574 is not divisible by 4.
It is because the number formed by its tens and the ones digits is 74, which is not divisible by 4.
(vi) 12030624 by 8
12030624 is divisible by 8.
It is because the number formed by its hundreds, tens and ones digits is 624, which is divisible by 8.

#### Question 15:

Which of the following are prime numbers?
(i) 103
(ii) 137
(iii) 161
(iv) 179
(v) 217
(vi) 277
(vii) 331
(viii) 397

A number between 100 and 200 is a prime number if it is not divisible by any prime number less than 15.
Similarly, a number between 200 and 300 is a prime number if it is not divisible by any prime number less than 20.

(i) 103 is a prime number, because it is not divisible by 2, 3, 5, 7, 11 and 13.
(ii) 137 is a prime number, because it is not divisible by 2, 3, 5, 7 and 11.
(iii) 161 is a not prime number, because it is divisible by 7.
(iv) 179 is a prime number, because it is not divisible by 2, 3, 5, 7, 11 and 13.
(v) 217 is a not prime number, because it is divisible by 7.
(vi) 277 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.
(vii) 331 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.
(viii) 397 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.

#### Question 16:

Give an example of a number
(i) which is divisible by 2 but not by 4.
(ii) which is divisible by 4 but not by 8.
(iii) which is divisible by both 2 and 8 but not by 16.
(iii) which is divisible by both 3 and 6 but not by 18.

(i) 14 is divisible by 2, but not by 4.
(ii) 12 is divisible by 4, but not by 8.
(iii) 24 is divisible by both 2 and 8, but not by 16.
(iv) 30 is divisible by both 3 and 6, but not by 18.

#### Question 17:

Write (T) for true and (F) for false against each of the following statements:
(i) If a number is divisible by 4, it must be divisible by 8.
(ii) If a number is divisible by 8, it must be divisible by 4.
(iii) If a number divides the sum of two numbers exactly, it must exactly divide the numbers separately.
(iv) If a number is divisible by both 9 and 10, it must be divisible by 90.
(v) A number is divisible by 18 if it is divisible by both 3 and 6.
(vi) If a number is divisible by 3 and 7, it must be divisible by 21.
(vii) The sum of two consecutive odd numbers is always divisible by 4.
(viii) If a number divides two numbers exactly, it must divide their sum exactly.

(i) If a number is divisible by 4, it must be divisible by 8. False
Example: 28 is divisible by 4 but not divisible by 8.

(ii) If a number is divisible by 8, it must be divisible by 4. True
Example: 32 is divisible by both 8 and 4.

(iii) If a number divides the sum of two numbers exactly, it must exactly divide the numbers separately. False
Example: 91 (51 + 40) is exactly divisible by 13. However, 13 does not exactly divide 51 and 40.

(iv) If a number is divisible by both 9 and 10, it must be divisible by 90. True
Example: 900 is both divisible by 9 and 10. It is also divisible by 90.

(v) A number is divisible by 18 if it is divisible by both 3 and 6. False
A number has to be divisible by 9 and 2 to be divisible by 18.
Example: 48 is divisible by 3 and 6, but not by 18.

(vi) If a number is divisible by 3 and 7, it must be divisible by 21. True
Example: 42 is divisible by both 3 and 7. It is also divisible by 21.

(vii) The sum of consecutive odd numbers is always divisible by 4. True
Example: 11 and 13 are consecutive odd numbers.
11 + 13 = 24, which is divisible by 4.

(viii) If a number divides two numbers exactly, it must divide their sum exactly. True
Example: 42 and 56 are exactly divisible by 7.
42+56 = 98, which is exactly divisible by 7.

#### Question 1:

Give the prime factorization of each of the following numbers:
12

We use the division method as shown below:

#### Question 2:

Give the prime factorization of each of the following numbers:
18

#### Question 3:

Give the prime factorization of each of the following numbers:
48

We will use the division method as shown below:

#### Question 4:

Give the prime factorization of each of the following numbers:
56

We will use the division method as shown below:

#### Question 5:

Give the prime factorization of each of the following numbers:
90

We will use the division method as shown below:

#### Question 6:

Give the prime factorization of each of the following numbers:
136

We will use the division method as shown below:

#### Question 7:

Give the prime factorization of each of the following numbers:
252

We will use the division method as shown below:

#### Question 8:

Give the prime factorization of each of the following numbers:
420

We will use the division method as shown below:

#### Question 9:

Give the prime factorization of each of the following numbers:
637

We will use the division method as shown below:

#### Question 10:

Give the prime factorization of each of the following numbers:
945

We will use the division method as shown below:

#### Question 11:

Give the prime factorization of each of the following numbers:
1224

We will use the division method as shown below:

#### Question 12:

Give the prime factorization of each of the following numbers:
1323

We will use the division method as shown below:

#### Question 13:

Give the prime factorization of each of the following numbers:
8712

We will use the division method as shown below:

#### Question 14:

Give the prime factorization of each of the following numbers:
9317

We will use the division method as shown below:

#### Question 15:

Give the prime factorization of each of the following numbers:
1035

We will use the division method as shown below:

#### Question 16:

Give the prime factorization of each of the following numbers:
1197

We will use the division method as shown below:

#### Question 17:

Give the prime factorization of each of the following numbers:
4641

We will use the division method as shown below:

#### Question 18:

Give the prime factorization of each of the following numbers:
4335

We will use the division method as shown below:

#### Question 19:

Give the prime factorization of each of the following numbers:
2907

We will use the division method as shown below:

#### Question 20:

Give the prime factorization of each of the following numbers:
13915

We will use the division method as shown below:

#### Question 1:

Find the HCF of the numbers in each of the following, using the prime factorization method:
84, 98

The given numbers are 84 and 98.

84 = 2 × 2 × 3 × 7 = 2 2 × 3 × 7
98 = 2 × 7 × 7 = 2 × 7 2

&there4 HCF of the given numbers = 2 × 7 = 14

#### Question 2:

Find the HCF of the numbers in each of the following, using the prime factorization method:
170, 238

The given numbers are 170 and 238.

&there4 H.C.F. of the given numbers = 2 × 17 = 34

#### Question 3:

Find the HCF of the numbers in each of the following, using the prime factorization method:
504, 980

The given numbers are 504 and 980.

504 = 2 × 2 ×2 × 3 × 3 × 7 = 2 3 × 3 2 × 7
980 = 2 × 2 × 5× 7 × 7 = 2 2 × 5 × 7 2
&there4 HCF of the given numbers = 2 2 × 7 = 28

#### Question 4:

Find the HCF of the numbers in each of the following, using the prime factorization method:
72, 108, 180

The given numbers are 72, 108 and 180

#### Question 5:

Find the HCF of the numbers in each of the following, using the prime factorization method:
84, 120, 138

The given numbers are 84, 120 and 138.

#### Question 6:

Find the HCF of the numbers in each of the following, using the prime factorization method:
106, 159, 371

The given numbers are 106, 159 and 371.
We have:

#### Question 7:

Find the HCF of the numbers in each of the following, using the prime factorization method:
272, 425

Given numbers are 272 and 425.

#### Question 8:

Find the HCF of the numbers in each of the following, using the prime factorization method:
144, 252, 630

The given numbers are 144, 252 and 630.
We have:

#### Question 9:

Find the HCF of the numbers in each of the following, using the prime factorization method:
1197, 5320, 4389

The given numbers are 1197, 5320 and 4389.

#### Question 10:

Find the HCF of the numbers in each of the following, using the division method:
58, 70

&there4 The HCF of 58 and 70 is 2.

#### Question 11:

Find the HCF of the numbers in each of the following, using the division method:
399, 437

The given numbers are 399 and 437.

#### Question 12:

Find the HCF of the numbers in each of the following, using the division method:
1045, 1520

The given numbers are 1045 and 1520.
We have:

&there4 The HCF of 1045 and 1520 is 95.

#### Question 13:

Find the HCF of the numbers in each of the following, using the division method:
1965, 2096

The given numbers are 1965 and 2096.

#### Question 14:

Find the HCF of the numbers in each of the following, using the division method:
2241, 2324

The given numbers are 2241and 2341.
We have:

#### Question 15:

Find the HCF of the numbers in each of the following, using the division method:
658, 940, 1128

The given numbers are 658, 940 and 1128.

First we will find the HCF of 658 and 940.

Thus, the HCF of 658 and 940 is 94.

Now , we will find the HCF of 94 and 1128.

Thus, the HCF of 94 and 1128 is 94.

&there4 The HCF of 658, 940 and 1128 is 94.

#### Question 16:

Find the HCF of the numbers in each of the following, using the division method:
754, 1508, 1972

The given numbers are 754, 1508 and 1972.

First, we will find the HCF of 754 and 1508.

So, the HCF of 754 and 1508 is 754.

Now, we will find the HCF of 754 and 1972.

So, the HCF of 754 and 1972 is 58.

&there4 The HCF of 754, 1058 and 1972 is 58.

#### Question 17:

Find the HCF of the numbers in each of the following, using the division method:
391, 425, 527

The given numbers are 391, 425 and 527.
First, we will find the HCF of 391 and 425.

So, the HCF of 391 and 425 is 17.
Now, we will find the HCF of 17 and 527.

So, the HCF of 17 and 527 is 17.
&there4 The HCF of 391, 425 and 527 is 17.

#### Question 18:

Find the HCF of the numbers in each of the following, using the division method:
1794, 2346, 4761

The given numbers are 1794, 2346 and 4761.
First, we will find the HCF of 1794 and 2346.

So, the HCF of 1794 and 2346 is 138.
Now, we will find the HCF of 138 and 4761.

So, the HCF of 138 and 4761 is 69.

&there4 The HCF of 1794, 2346 and 4761 is 69.

#### Question 19:

Show that the following pairs are co-primes:
59, 97

The given numbers are 59 and 97.

Since 59 and 97 does not have any common factor other than 1, the two numbers are co-primes.

#### Question 20:

Show that the following pairs are co-primes:
161, 192

Now, 161 = 7 × 23 × 1
192 = 2 × 2× 2 ×2 × 2× 2 × 3 = 2 6 × 3 × 1
&there4 HCF = 1
Hence, 161 and 192 are co-primes.

#### Question 21:

Show that the following pairs are co-primes:
343, 432

Now, 343 = 7 × 7× 7 × 1 = 7 3 × 1
432 = 2 × 2× 2 ×2 × 3× 3 ×3 = 2 4 × 3 3 × 1
&there4 HCF =1
Hence, 343 and 432 are co-primes.

#### Question 22:

Show that the following pairs are co-primes:
512, 945

512 = 2 × 2 ×2 × 2 × 2× 2 × 2× 2 × 2 = 2 9
945 = 3 × 3 × 3 × 5 × 7 = 3 3 × 5 × 7
Thus, the HCF of 512 and 945 is 1.
&there4 512 and 945 are co-primes.

#### Question 23:

Show that the following pairs are co-primes:
385, 621

#### Question 24:

Show that the following pairs are co-primes:
847, 1014

The given numbers are 847 and 1014.

847 = 7 × 11 × 11 × 1 = 7 × 11 2 × 1
1014 = 2 × 3 × 13 × 13 × 1
&there4 HCF = 1
Hence, 847 and 1014 are co-primes.

#### Question 25:

Find the greatest number which divides 615 and 963, leaving the remainder 6 in each case.

Because the remainder is 6, we have to find the number that exactly divides (615 - 6) and (963 - 6).

Therefore, the required number is 87.

#### Question 26:

Find the greatest number which divides 2011 and 2623, leaving remainders 9 and 5 respectively.

&there4 The required number is 154 .

#### Question 27:

Find the greatest number that will divide 445, 572 and 699, leaving remainders 4, 5, 6 respectively.

Since the respective remainders of 445, 572 and 699 are 4, 5 and 6, we have to find the number which exactly divides (445-4), (572-5) and (696-6).

So, the required number is the HCF of 441, 567 and 693.
Firstly, we will find the HCF of 441 and 567.

Now, we will find the HCF of 63 and 693 .

&there4 HCF = 63

Hence, the required number is 63 .

#### Question 28:

Reduce each of the following fractions to the lowest terms:
(i) 161 207
(ii) 517 799
(iii) 296 481

(i) 161 207
To reduce the given fraction to its lowest term, we will divide the numerator and the denominator by their HCF.

Now, we will find the HCF of 161 and 207 .

Dividing the numerator and the denominator by the HCF, we get:

(ii) 517 799
To reduce the given fraction to its lowest term, we will divide the numerator and the denominator by their HCF.
Now, we will find the HCF of 517 and 799.

Dividing the numerator and the denominator by the HCF, we get:

(iii) 296 481
To reduce the given fraction to its lowest term, we will divide the numerator and the denominator by their HCF.
Now, we will find the HCF of 296 and 481.

Dividing the numerator and the denominator by the HCF, we get:

#### Question 29:

Three pieces of timber, 42-m, 49-m and 63-m long, have to be divided into planks of the same length. What is the greatest possible length of each plank?

The lengths of the three pieces of timber are 42 m, 49 m and 63 m.
The greatest possible length of each plank will be given by the HCF of 42, 49 and 63 .

Firstly, we will find the HCF of 42 and 49 by division method.

&there4 The HCF of 42 and 49 is 7.
Now, we will find the HCF of 7 and 63.
​​

&there4 The HCF of 7 and 63 is 7.
Therefore, HCF of all three numbers is 7
Hence, the greatest possible length of each plank is 7 m .

#### Question 30:

Three different containers contain 403 L, 434 L and 465 L of milk respectively. Find the capacity of a container which can measure the milk of all the containers in an exact number of times.

Three different containers contain 403 L, 434 L and 465 L of milk.

The capacity of the container that can measure the milk in an exact number of times will be given by the HCF of 403, 434 and 465 .

Now, we will find the HCF of 31 and 465 .

Hence, the capacity of the required container is 31 L.

#### Question 31:

There are 527 apples, 646 pears and 748 oranges. These are to be arranged in heaps containing the same number of fruits. Find the greatest number of fruits possible in each heap. How many heaps are formed?

Number of apples = 527
Number of pears = 646
Number of oranges = 748
The fruits are to be arranged in heaps containing the same number of fruits.
The greatest number of fruits possible in each heap will be given by the HCF of 527, 646 and 748 .

Firstly, we will find the HCF of 527 and 646 .

&there4 H CF of 527, 646 and 748 = 17

So, the greatest number of fruits in each heap will be 17 .

#### Question 32:

Determine the longest tape which can be used to measure exactly the lengths 7 m, 3 m 85 cm and 12 m 95 cm.

7 m = 700 cm
3 m 85 cm = 385 cm
12 m 95 cm = 1295 cm

The required length of the tape that can measure the lengths 700 cm, 385 cm and 1295 cm will be given bu the HCF of 700 cm, 385 cm and 1295 cm .

Evaluating the HCF of 700, 385 and 1295 using prime factorisation method, we have:

Hence, the longest tape which can measure the lengths 7 m, 3 m 85 cm and 12 m 95 cm exactly is of 35 cm.

#### Question 33:

A rectangular courtyard is 18 m 72 cm long and 13 m 20 cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.

Length of the courtyard = 18 m 72 cm = 1872 cm
Breadth of the courtyard = 13 m 20 cm = 1320 cm

Now, maximum edge of the square tile is given by the HCF of 1872 cm and 1320 cm.

HCF of 1872 and 1320 = 24
&there4 maximum edge of the square tile = 24 cm

Required   number   of   tiles   =   area   of   courtyard area   of   each   square   tile = 1872 × 1320 24 × 24 = 4290

#### Question 34:

Find the HCF of
(i) two prime numbers
(ii) two consecutive numbers
(iii) two co-primes
(iv) 2 and an even number

(i) 2 and 3 are two prime numbers.
Now, HCF of 2 and 3 is as follows:
2 = 2 × 1
3 = 3 × 1
&there4 HCF = 1

(ii) 4 and 5 are t wo consecutive numbers.
Now, HCF of 4 and 5 is as follows :
4 = 2 × 2 × 1 = 2 2 × 1
5 = 5 × 1
&there4 HCF = 1

(iii) 2 and 3 are t wo co-primes.
Now, HCF of 2 and 3 is as follows :
2 = 2 × 1
3 = 3 × 1
&there4 HCF = 1

(iv) 2 and 4 are two even numbers.
Now, HCF of 2 and 4 is as follows:​
2 = 2 × 1
4 = 2 × 2 × 1
&there4 HCF = 2 × 1 = 2

#### Question 1:

Find the LCM of the numbers given below:
42, 63

The given numbers are 42 and 63.

#### Question 2:

Find the LCM of the numbers given below:
60, 75

The given numbers are 60 and 75.

#### Question 3:

Find the LCM of the numbers given below:
12, 18, 20

The given numbers are 12, 18 and 20.
We have:

#### Question 4:

Find the LCM of the numbers given below:
36, 60, 72

The given numbers are 36, 60 and 72.

#### Question 5:

Find the LCM of the numbers given below:
36, 40, 126

The given numbers are 36, 40 and 126.

#### Question 6:

Find the LCM of the numbers given below:
16, 28, 40, 77

The given numbers are 16, 28, 40 and 77.
We have:

#### Question 7:

Find the LCM of the numbers given below:
28, 36, 45, 60

The given numbers are 28, 36, 45 and 60.

#### Question 8:

Find the LCM of the numbers given below:
144, 180, 384

#### Question 9:

Find the LCM of the numbers given below:
48, 64, 72, 96, 108

The given numbers are 48, 64, 72, 96 and 108.
We have:

#### Question 10:

Find the HCF and LCM of
117, 221

The given numbers are 117 and 221.

Now,
117 = 3 × 3 × 13
221 = 13 × 17

Now, LCM = 13 × 17 × 3 × 3
= 1989

#### Question 11:

Find the HCF and LCM of
234, 572

The given numbers are 234 and 572.

&there4 LCM = 13 × 2 × 2 × 11 × 9
= 5148
Also, HCF = 13 × 2 = 26

#### Question 12:

Find the HCF and LCM of
693, 1078

The given numbers are 693 and 1078.

#### Question 13:

Find the HCF and LCM of
145, 232

&there4 HCF = 29
Also, LCM = 29 × 2 × 2 × 2 × 5 = 1160

#### Question 14:

Find the HCF and LCM of
861, 1353

The given numbers are 861 and 1353.

861 = 3 × 41 × 7
1353 = 41 × 11 × 3

&there4 HCF = 41 × 3 = 123
Also, LCM = 41 × 3 × 11 × 7 = 9471

#### Question 15:

Find the HCF and LCM of
2923, 3239

We know that product of two numbers = HCF × LCM

#### Question 16:

For each pair of numbers, verify that their product = (HCF × LCM).
(i) 87, 145
(ii) 186, 403
(iii) 490, 1155

We have:
87 = 3 × 29
145 = 5 × 29

HCF = 29
LCM = 29 × 15 × 1 = 435

Now, HCF × LCM = 29 × 435 = 12615
Product of the two numbers = 87 × 145 = 12615

&there4 HCF × LCM = Product of the two numbers
Verified.

(ii)186 and 403

186 = 2 × 3 × 31
403 = 31 × 13

HCF = 31
LCM = 31 × 13 × 6 = 2418

Now, HCF × LCM = 31 × 2418 = 74958
Product of the two numbers = 186 × 403 = 74958

&there4 HCF × LCM = Product of the two numbers
Verified.

HCF = 7 × 5 = 35
LCM = 7 × 5 ×7 × 2 × 3 × 11 = 16170

Now, HCF × LCM = 35 × 16170 = 565950
Product of the two numbers = 490 × 1155 = 565950

&there4 HCF × LCM = Product of the two numbers
Verified.

#### Question 17:

The product of two numbers is 2160 and their HCF is 12. Find their LCM.

Product of the two numbers = 2160
HCF = 12

We know that LCM × HCF = Product of the two numbers

&there4 LCM = 2160 12 = 180

#### Question 18:

The product of two numbers is 2160 and their LCM is 320. Find their HCF.

Product of the two numbers = 2560
LCM = 320

LCM × HCF = Product of the two numbers

#### Question 19:

The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find the other.

HCF = 145
LCM = 2175
One of the number = 725

We know that
HCF × LCM = Product of two numbers
&there4 Other number = 145 × 2175 725 = 435

#### Question 20:

The HCF and LCM of two numbers are 131 and 8253 respectively. If one of the numbers is 917, find the other.

HCF = 131
LCM = 8253
One of the number = 917

We know that
LCM × HCF = Product of two numbers
Other number = 8253 × 131 917

#### Question 21:

Find the least number divisible by 15, 20, 24, 32 and 36.

The given numbers are 15, 20, 24, 32 and 36.

The smallest number divisible by the numbers given above will be their LCM.

LCM = 2 5 × 3 2 × 5
= 1440
&there4 The least number divisible by 15, 20, 24, 32 and 36 is 1440 .

#### Question 22:

Find the least number which when divided by 25, 40 and 60 leaves 9 as the remainder in each case.

25, 40 and 60 exactly divides the least number that is equal to their LCM.
So, the required number that leaves 9 as a remainder will be LCM + 9.

#### Question 23:

Find the least number of five digits that is exactly divisible by 16, 18, 24 and 30.

We have to find the least five-digit number that is exactly divisible by 16, 18, 24 and 30.
But LCM=720 is a three digit number.

#### Question 24:

Find the greatest number of five digits exactly divisible by 9, 12, 15, 18 and 24.

#### Question 25:

Three bells toll at intervals of 9, 12, 15 minutes. If they start tolling together, after what time will they next toll together?

Three bells toll at intervals of 9, 12 and, 15 minutes.
The time when they will toll together again is given by the LCM of 9, 12 and 15.

Required time = 2 2 × 3 2 × 5
= 180 minutes
=3 h
If they start tolling together, they will toll together again after 3 h.

#### Question 26:

Three boys step off together from the same place. If their steps measure 36 cm, 48 cm and 54 cm, at what distance from the starting point will they again step together?

From the starting point, they will step together again when they travel a distance that is exactly divisible by the lengths of their steps.
The least distance from the starting point where they will step together will be given by the LCM of 36, 48 and 54.

2 36 , 48 , 54 2 18 , 24 , 27 3 9 , 12 , 27 3 3 , 4 , 9 3 1 , 4 , 3 2 1 , 4 , 1 ق 1 , 2 , 1     ف,1,1

The required distance = 2 × 2 ×3 × 3 × 3 × 2 × 2
= 16 × 27
= 432 cm
&there4 They will step together again at a distance of 432 cm from the starting point.

#### Question 27:

The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds. If they start changing simultaneously at 8 a.m., after how much time will they change again simultaneously?

The time when the lights will change simultaneously again will be quantity which is exactly divisible by 48, 72 and 108. The least time when they change simultaneously will be given by their LCM.

2 48 , 72 , 108 2 24 , 36 , 54 2 12 , 18 , 27 2 6 , 9 , 27 3 3 , 9 , 27 3 1 , 3 , 9 3 1 , 1 , 3  ف,1,1
Required time = 2 4 × 3 3
= 432 seconds
= 7 min 12 seconds
So, the lights will change simultaneously at 8:07:12 a.m.

#### Question 28:

Three measuring rods are 45 cm, 50 cm and 75 cm in length. What is the least length (in metres) of a rope that can be measured by the full length of each of these three rods?

The length of the required rope must be such that it is exactly divisible by 45, 50 and 75. The least length will be given by the LCM of 45, 50 and 75.

2 45 , 50 , 75 3 45 , 25 , 75 3 15 , 25 , 25 5 5 , 25 , 25 5 1 , 5 , 5   ف,1,1

Required length = 3 × 3 ×5 × 5 × 2
= 450 cm
So, the minimum length of the rope that can be measured by the full length of each of the three rods is 450 cm.

#### Question 29:

An electronic device makes a beep after every 15 minutes. Another device makes a beep after every 20 minutes. They beeped together at 6 a.m. At what time will they next beep together?

The LCM of the time intervals of the beeps will give the time when the electronic devices will beep together.

5 15 , 20 3 3 , 4 2 1 , 4 2 1 , 2       1 , 1

Required time = 5 × 3 × 2 × 2
= 60 min
So, they will beep simultaneously after 60 min or 1 h.

&there4 They will beep together again at 7:00 a.m .

#### Question 30:

The circumferences of four wheels are 50 cm, 60 cm, 75 cm and 100 cm. They start moving simultaneously. What least distance should they cover so that each wheel makes a complete number of revolutions?

Distance covered by a wheel for one complete revolution = circumference of the wheel

All the wheels will make complete numbers of revolutions when the distances covered by them is equal to their LCM.
5 50 , 60 , 75 , 100 5 10 , 12 , 15 , 20 2 2 , 12 , 3 , 4 2 1 , 6 , 3 , 2 3 1 , 3 , 3 , 1   ف,1,1,1
Required least distance = 5 × 5 ×2 × 2 × 3
= 25 × 4 × 3
= 300 cm = 3 m
So, each wheel will make a complete number of revolutions after travelling 3 m.

#### Question 1:

Which of the following numbers is divisible by 3?
(a) 24357806
(b) 35769812
(c) 83479560
(d) 3336433

A number is divisible by 3 if the sum of its digits is divisible by 3.

a) Consider the number 24357806.
Sum of its digits = 2 + 4 + 3 + 5+ 7 + 8 + 0 + 6 = 35, which is not divisible by 3.
So, 2357806 is not divisible by 3.

b) Consider the number 35769812.
Sum of its digits = 3 + 5 + 7 + 6 +9 + 8 + 1 + 2 = 41, which is not divisible by 3.
So, 35769812 is not divisible by 3.

c) Consider the number 83479560.
Sum of its digits = 8 + 3 + 4+ 7 + 9 + 5 + 6 + 0 = 42, which is divisible by 3.
So, 2357806 is divisible by 3.

d) Consider the number 3336433.
Sum of its digits = 3 + 3 +3 + 6 +4 + 3 +3 = 25, which is not divisible by 3.
So, 3336433 is not divisible by 3.

#### Question 2:

Which of the following numbers is divisible by 9?
(a) 8576901
(b) 96345210
(c) 67594310
(d) none of these

A number is divisible by 9 if the sum of its digits is divisible by 9.

a) Consider the number 8576901.
Sum of its digits = 8 + 5 +7 + 6 + 9+ 0 + 1 = 36, which is divisible by 9.
So, 8576901 is divisible by 9.

b) Consider the number 96345210.
Sum of its digits = 9 + 6 + 3+ 4 + 5+ 2 + 1 + 0 = 30, which is not divisible by 9.
So, 96345210 is not divisible by 9.

c) Consider the number 67594310.
Sum of its digits = 6 + 7 + 5 + 9 + 4 + 3 + 1 + 0 = 35, which is not divisible by 9.
So, 67594310 is not divisible by 9.

#### Question 3:

Which of the following numbers is divisible by 4?
(a) 78653234
(b) 98765042
(c) 24689602
(d) 87941032

A number is divisible by 4 if the number formed by its digits in the tens and ones places is divisible by 4.

(a) 78653234
Consider the number 78653234.
Here, the number formed by the tens and the ones digit is 34, which is not divisible by 4.
Therefore, 78653234 is not divisible by 4.

(b) 98765042
Consider the number 98765042.
Here, the number formed by the tens and the ones digit is 42, which is not divisible by 4.
Therefore, 98765042 is not divisible by 4.

(c) 24689602
Consider the number 24689602.
Here, the number formed by the tens and the ones digit is 02, which is not divisible by 4.
Therefore, 24689602 is not divisible by 4

(d) 87941032
Consider the number 87941032.
Here, the number formed by the tens and ones digit is 32, which is divisible by 4.
Therefore, 87941032 is divisible by 4.

#### Question 4:

Which of the following numbers is divisible by 8?
(a) 96354142
(b) 37450176
(c) 57064214
(d) none of these

A number is divisible by 8 if the number formed by its digits in hundreds, tens and ones places is divisible by 8.

(a) 96354142
Consider the number 96354142.
Here, the number formed by the digits in hundreds, tens and ones places is 142, which is clearly not divisible by 8.
Therefore, 96354142 is not divisible by 8.

(b) 37450176
Consider the number 37450176.
The number formed by the digits in hundreds, tens and ones places is 176, which is clearly divisible by 8.
Therefore, 37450176 is divisible by 8.

(c) 57064214
Consider the number 57064214.
Here, the number formed by the digits in hundreds, tens and ones places is 214, which is clearly not divisible by 8.
Therefore, 57064214 is not divisible by 8.

#### Question 5:

Which of the following numbers is divisible by 6?
(a) 8790432
(b) 98671402
(c) 85492014
(d) none of these

A number is divisible by 6, if it is divisible by both 2 and 3.

(a) 8790432
Consider the number 8790432.
The number in the ones digit is 2.
Therefore, 8790432 is divisible by 2.
Now, the sum of its digits (8+7+9+0+2+3+2) is 33. Since 33 is divisible by 3, we can say that 8790432 is also divisible by 3.
Since 8790432 is divisible by both 2 and 3, it is also divisible by 6.

(b) 98671402
Consider the number 98671402.
The number in the ones digit is 2.
Therefore, 98671402 is divisible by 2.
Now, the sum of its digits (9+8+6+7+1+4+0+2) is 37. Since 37 is not divisible by 3, we can say that 98671402 is also not divisible by 3.
Since 98671402 is not divisible by both 2 and 3, it is not divisible by 6.

(c) 85492014
Consider the number 85492014.
The number in the ones digit is 4.
Therefore, 85492014 is divisible by 2.
Now, the sum of its digits (8+5+4+9+2+0+1+4) is 33. Since 33 is divisible by 3, we can say that 85492014 is also divisible by 3.
Since 85492014 is divisible by both 2 and 3, it is also divisible by 6.

#### Question 6:

Which of the following numbers is divisible by 11?
(a) 3333333
(b) 1111111
(c) 22222222
(d) none of these

(c) 22222222
A number is divisible by 11, if the difference of the sum of its digits in odd places and the sum of the digits in even places (starting from ones place) is either 0 or a multiple of 11.

(a) 3333333
Consider the number 3333333.
Sum of its digits in odd places (3 + 3 + 3 + 3) = 12
Sum of its digits in even places (3 + 3 + 3) = 9
Difference of the two sums = 12 &minus 9 = 3
Since this number (3) is not divisible by 11, 3333333 is not divisible by 11.

(b) 1111111
Consider the number 1111111.
Sum of its digits in odd places (1 + 1 + 1 + 1) = 4
Sum of its digits in even places (1 + 1 + 1) = 3
Difference of the two sums = 4 &minus 3 = 1
Since this number (1) is not divisible by 11, 1111111 is also not divisible by 11.

(c) 22222222
Consider the number 22222222.
Sum of its digits in odd places (2 + 2 + 2 + 2)= 8
Sum of its digits in even places (2 + 2 + 2 + 2) = 8
Difference of the two sums = 8 &minus 8 = 0
Since this number (0) is divisible by 11, 22222222 is also divisible by 11.

#### Question 7:

Which of the following is a prime number?
(a) 81
(b) 87
(c) 91
(d) 97

(a) 81 is not a prime number because 81 can be written as 9×9.
(b) 87 is not a prime number because 87 can be written as 29×3.
(c) 91 is not a prime number because 91 can be written as 13×7.
(d) 97 is a prime number.

#### Question 8:

Which of the following is a prime number?
(a) 117
(b) 171
(c) 179
(d) none of these

(a) 117 is not a prime number because 117 can be written as 3 × 39.
(b) 171 is not a prime number because 171 can be written as 19×9.
(c) 179 is prime number.

#### Question 9:

Which of the following is a prime number?
(a) 323
(b) 361
(c) 263
(d) none of these

(a) 323 is not a prime number because 323 can be written as 17 × 19.
(b) 361 is not a prime number because 361 can be written as 19 × 19.
(c) 263 is a prime number.

#### Question 10:

Which of the following are co-primes?
(a) 8, 12
(b) 9, 10
(c) 6, 8
(d) 15, 18

(a) 8, 12 are not co-primes as they have a common factor 4.
(b) 9, 10 are co-primes as they do not have a common factor.
(c) 6, 8 are not co-primes as they have a common factor 2.
(d)15,18 are not co-primes as they have a common factor 3.

#### Question 11:

Which of the following is a composite number?
(a) 23
(b) 29
(c) 32
(d) none of these

(a) 23 is not a composite number as it cannot be broken into factors.
(b) 29 is not a composite number as it cannot be broken into factors.
(c) 32 is a composite number as it can be broken into factors, which are 2 × 2 × 2 × 2 × 2.

#### Question 12:

The HCF of 144 and 198 is
(a) 9
(b) 12
(c) 6
(d) 18

(d) 2 × 3 2 = 18
We first factorise the two numbers:

144 = 2 × 2 × 2 × 2 × 3 × 3 = 2 4 × 3 2
198 = 2 × 3 × 3 × 11 = 2 × 3 2 × 11
Here, 18 (2 × 3 2 = 18) is the highest common factor of the two numbers.

#### Question 13:

The HCF of 144, 180 and 192 is
(a) 12
(b) 16
(c) 18
(d) 8

(a) 2 2 ×3= 12
We will first factorise the two numbers:

Here, 12 (i.e. 2 2 × 3 = 12) is the highest common factor of the three numbers.

#### Question 14:

Which of the following are co-primes?
(a) 39, 91
(b) 161, 192
(c) 385, 462
(d) none of these

(a) 39 and 91 are not co-primes as 39 and 91 have a common factor, i.e. 13.
(b) 161 and 192 are co-primes as 161 and 192 have no common factor other than 1.
(c) 385 and 462 are not co-primes as 385 and 462 have common factors 7 and 11 .

#### Question 15:

289 391 when reduced to the lowest terms is
(a) 11 23
(b) 13 31
(c) 17 31
(d) 17 23

Dividing both the numerator and the denominator by the H.C.F. of 289 & 391:

#### Question 16:

The greatest number which divides 134 and 167 leaving 2 as remainder in each case is
(a) 14
(b) 17
(c) 19
(d) 11

(d) 11
Since we need 2 as the remainder, we will subtract 2 from each of the numbers.
167 &minus 2 = 165
134 &minus 2 = 132
Now, any of the common factors of 165 and 132 will be the required divisor.
On factorising:
165 = 3 × 5 × 11
132 = 2 × 2 × 3 × 11
Their common factors are 11 and 3.
So, 11 is the required divisor.

#### Question 17:

The LCM of 24, 36, 40 is
(a) 4
(b) 90
(c) 360
(d) 720

## Rule 9: Factoring Integers

To factor an integer, simply break the integer down into a group of numbers whose product equals the original number. Factors are separated by multiplication signs. Note that the number 1 is the factor of every number. All factors of a number can be divided evenly into that number.

Example 1: Factor the number 3.

Answer: Since 3 x 1 = 3, the factors of 3 are 3 and 1.

Example 2: Factor the number 10.

Answer: Since 10 can be written as 5 x 2 x 1, the factors of 10 are 10, 5, 2, and 1. The number 10 can be divided by 10, 5, 2, and 1.

Example 3: Factor the number 18.

Answer: The number 18 can be written as 18 x 1 or 9 x 2 or 6 x 3 or 3 x 3 x 2. Since 18 can be divided by 18, 9, 6, 3, 2, and 1, then 18, 9, 6, 3, 2, and 1 are factors of 18.

Example 4: Factor the number 24.

Answer: The number 24 can be written as 24 x 1 or 12 x 2 or 8 x 3 or 4 x 6 or 2 x 2 x 2 x 3. Since 24 can be divided by 24, 12, 8, 6, 4, 3, 2, and 1, then 24, 12, 8, 6, 4, 3, 2, and 1 are factors of 24.

Example 5: Factor the number 105.

Answer: The number 105 can be written as 105 x 1 or 21 x 5 or 3 x 7 x 5 or 15 x 7 or 35 x 3. Since 105 can be divided by 105, 35, 21, 15, 7, 5, 3, and 1, then 105, 35, 21, 15, 7, 5, 3, and 1 are factors of 105.

Example 6: Factor the number 1200 completely.

Answer: This instruction means to factor 1200 into a set of prime factors (factors that cannot again be factored). The number 1200 can be written as 1200 x 1 or 100 x 12. Note the 100 can again be factored to 10 x 10 and the 12 can be factored to 6 x 2. So now you have 1200 = 100 x 12 = 10 x 10 x 6 x 2. This factored set can again be factored to (2 x 5) x (2 x 5) x (2 x 3) x 2 x 1. The number 1200 is factored completely as 5 x 5 x 3 x 2 x 2 x 2 x 2 x 1.