Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers

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Kerala State Syllabus 6th Standard Maths Solutions Chapter 6 Numbers

Numbers Text Book Questions and Answers

Let’s make a rectangle Textbook Page No. 95

A rectangle with 20 dots:
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 1
5 dots wide, 4 dots high.
Can we make other rectangles, rearranging the dots?
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 4
How about this?
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 2
Also like this:
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 3
Any other?
What about the number of dots along the width and height?
Their product must be 20, right?
In what all ways can we write 20 as the product of two natural numbers?
Answer:
4 and 5 are the two natural numbers.
Explanation:
4 x 5 = 20
5 x 4 = 20

Now make different rectangles with 24 dots. Also write down the number of dots along the width and height.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 5
Answer:

Explanation:
Total number of 24 dots,
2, 4, 6 and 12 are factors of 24,
and the product of 4 x 6 = 24 or 6 x 4 =24
or 2 x 12  or 12 x 2 = 24

What about 30 dots?

Let’s think about it, without actually making rectangles. What are the possible number of dots along the width and height?
The product of numbers in every row of the table is 30.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 6
Answer:

Explanation:
5 x 6 = 30 or 6 x 5 = 30
5 dots horizontal by 6 dots vertical.
There’s another way of stating this; all these numbers are factors of 30.

Now can you write down the different rectangles with 40 dots?
How about 45 dots?
And 60 dots?
What about 61 dots?
Answer:

Explanation:
9 x 5 = 45
5 x 9 = 45
10 x 6 = 60
6 x 10= 60
61 not possible

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Factor pairs

What are the factors of 72?
Two quick ones are 1 and 72.
We can divide 72 by 2 without any remainder. That is 2 is also a factor of 72.And 72 divided by 2 gives 36.
72 = 2 × 36
So 36 is also a factor of 72.
Thus we can find factors in pairs.
Since
72 ÷ 3 = 24
We have
72 = 3 × 24
This gives 3 and 24 as another pair of factors.

Can’t we find other pairs like this?

(1, 72)

(2, 36)

(3, 24)

(4, 18)

(6, 12)

(8, 9)

Now try to find the factors of 90, 99, 120 as pairs.
Answer:
Factors of 90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.
Factors of 99 = 1, 3, 9, 11, 33, and 99.
Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
Explanation:
A factor is a number that divides the given number without any remainder.
The factors of a number can either be positive or negative.
factor pairs of 90
(1, 90)
(2, 45)
(3, 30)
(5, 18)
(6, 15)
(9, 10)
factor pairs of 99
(1, 99)
(3, 33)
(9, 11)
factor pairs of 120
(1, 120)
(2, 60)
(3, 40)
(4, 30)
(5, 24)
(6, 20)
(8, 15)
(10, 12)

• If 2 and 3 are factors of a number, should 6 also be a factor of that number’?
Answer:
yes,
Explanation:
Yes, 6 also be a factor of that number.
2 x 3 = 6
6 x 1 = 6
3 x 2 = 6

• If 3 and 5 are factors of a number, should 15 also be a factor of that number’?
Answer:
yes,
Explanation:
5 x 3 = 15
15 x 1 =15
3 x 5 = 15
3, 5 are the factors of 15.

• If 4 and 6 are factors of a number, should 24 also be a factor of that number’?
Answer:
Yes,
Explanation:
4 x 6 = 24
6 x 4 = 24
4 , 6 are the factors of 24

• If 4 and 6 are factors of a number, what is the largest number we can say for sure is a factor of that number’?
Answer:
12,
Explanation:
12 is the largest number is a factor of 24.
The largest number is 12.
The numbers whose factors and 4 and 6 are common multiples of 4 and 6 .
These numbers are also multiples of common multiple of 12.

• Given Two factors of a number, under what conditions can we say for sure that the product of these factors is also a factor’?
Answer:
2 and 12.
Explanation:
2 x 12 = 24
The numbers whose factors and 2 and 12 are common multiples of 2 and 12.
These numbers are also multiples of common multiple of 24.

Odd and even

We have found the factors of many numbers like 20, 24, 30, 40, 45, 60, 61, 72, 90, 99, 120.
See how many factors each has.
All of them have an even number of factors, right?
Why is this so?
Is it true for all numbers?
Write the factor pairs of 36.
(1, 36), (2, 18), (3, 12), (4, 9), (6, 6)
So what are the factors of 36?
1, 2, 3, 4, 6, 9, 12, 18, 36
9 factors in all.

Why is the number of factors odd in the case?
Can you find any other number with an odd number of factors?
Take 16, for example.
How about 25?
What is the specialty of numbers with an odd number of factors?
Answer:
1, 5, 25
Explanation:
1 x 25 = 25
5 x 5 =25
1, 5, 25 are the factors of 25.

Can you find all numbers between 1 and 100; which have an odd number of factors?
Answer:
Yes,
Explanation:
To find an odd factor, exclude the even prime factor 2.
The odd numbers from 1 to 100 are:
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Repeated multiplication Textbook Page No. 97

How many factors does 5 have?
How about 17?
5 and 17 are prime numbers, aren’t they?
And a prime has only two factors, right? 1 and the number itself.

All composite numbers have more than two factors.
For example, let’s have a look at 32.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 7
32 = 2 × 2 × 2 × 2 × 2
Taking the first 2 alone and all the other 2’s together, we can write
32 = 2 × 16
How about taking the first two 2 s together and then other 2’s together?
32 = 4 × 8
Taking all the 2’s together can be written as
32 = 1 × 32
Thus, the factors of 32 are the 6 numbers
1, 2, 4, 8, 16, 32
Let’s look at the factors of 81 like this:
Writing 81 as a product of prime numbers, we get
81 = 3 × 3 × 3 × 3
So we can write 81 as
3 × 27
9 × 9
1 × 81
Thus we have five factors.
1, 3, 9, 27, 81
We can put this in a different way.
Taking 3’s in groups we get the factors.
3
3 × 3 = 9
3 × 3 × 3 = 27
3 × 3 × 3 × 3 = 81
and find the 5 factors of 81 as 1, 3, 9, 27 and 81.
In these examples, 32 is a product of 2’s; and 81 is a product of 3’s.

Like this, can’t we easily find the factors of a number, which can be factorized as repeated product of a single prime?
Answer:
This is true as any composite number can be expressed in terms of prime factors.
Explanation:
The prime factor for number 2 is same for the numbers.
for example;
12 = 2 x 2 x 3
18 = 2 x 3 x 3
both have prime factors 2 and 3 but the combination 12 is different from that of 18.

We can split 216 as
216 = 6 × 6 × 6
Can we say that the only factors of 216 are the 4 numbers 1, 6, 36, 24. What are the other factors of 216?
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 16
Answer:
Therefore, the factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, and 216.
Explanation:
The factor of a number is a number that divides the given number without any remainder.
To find the factors of given number, divide the number with the least prime number, i.e. 2.
1 x 216 =216
2 x 108 = 216
3 x 72 = 216
4 x 54 = 216
6 x 36 = 216
8 x 27 = 216
9 x 24 = 216
12 x 18 = 216

Question 1.
Find all the factors of the numbers below:
(i) 256
Answer:
The factors of 256 are 1, 2, 4, 8, 16, 32, 64, 128 and 256.
Explanation:
The factors of 256 are 1, 2, 4, 8, 16, 32, 64, 128 and 256.
1 x 256 =256
2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2 x 128 = 256
2 x 2 x 64 = 4 x 64 = 256
2 x 2 x 2 x 64 = 8 x 32 = 256
2 x 2 x 2 x 2 x 16 = 16 x 16 = 256

(ii) 625
Answer:
The factors of 625 are 1, 5, 25, 125 and 625.
Explanation:
The factors of 625 are 1, 5, 25, 125 and 625.
1 x 625 = 625
5 x 125 = 625
25 x 25 = 625

(iii) 243
Answer:
The factors of 243 are 1, 3, 9, 27, 81 and 243.
Explanation:
The factors of 243 are 1, 3, 9, 27, 81 and 243.
3 x 81 = 243
9 x 27 = 243

(iv) 343
Answer:
The factors of 343 are 1, 7, 49 and 343.
Explanation:
The factors of 343 are 1, 7, 49 and 343.
1 x 343 = 343
7 x 49 = 343

(v) 121
Answer:
The factors of 121 are 1, 11 and 121.
Explanation:
The factors of 121 are 1, 11 and 121.
1 x 121 = 121
11 x 11 = 121

Question 2.
Which are the numbers between 1 and 100 having exactly three factors?
Answer:
4, 9 , 25 and 49.
Explanation:
1, 2 and 4 three factor for 4
1, 3 and 9 three factor for 9
1, 5 and 25 three factor for 25
1, 7 and 49 three factor for 49
We know that the numbers between 1 and 100,
which have exactly three factors are 4, 9, 25 and 49.

Prime factors Textbook Page No. 99

How do we find the factors of 16?
The only prime factor of 16 is 2. Writing
16 = 2 × 2 × 2 × 2
We see that the factors of 16, except 1, are products of 2’ s.
2
2 × 2 = 4
2 × 2 × 2 = 8
2 × 2 × 2 × 2 = 16
Taking 1 also, we get all the factors of 16 as 1,2, 4, 8, 16.
Now let’s try 16 × 3 = 48.
48 = (2 × 2 × 2 × 2) × 3
To get its factors, we can multiply some of the 2’s only; or some 2 ‘s and 3.
Taking only 2’s, what we get are the factors of 16.
2, 4, 8, 16
What if we take 2’s and 3?
(2 × 3) = 6
(2 × 2) × 3 = 4 × 3 = 12
(2 × 2 × 2) × 3 = 8 × 3 = 24
(2 × 2 × 2 × 2) × 3 = 48
Thus we get also the factors.
6, 12, 24, 48

3 alone is also a factor. Also 1,which is a factor of every number. We can separate these factors like this;
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 8
What is the relation between each number in the first row with the number below it.
Now let’s take 48 × 3 = 144
144 = (2 × 2 × 2 × 2) × (3 × 3)
The factors can be got by taking only some 2’ s, some 2’ s and one 3 or some 2’s and two 3’s.
Taking 3 ’s only we get 3 and 9.
And 1 also is a factor.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 9
These can also be written in a table like this:
The numbers in the first row, multiplied by 3, give the numbers in the second row.
And numbers in the second row, multiplied by 3, give the numbers in the third row.

Let’s look at the table along the columns.
First column is 1, 3, 9; these numbers do not have 2 as a factor.
Second column is 2, 6, 18; these have a single 2 as a factor.
What about the third and fourth columns?
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 10
Thus the numbers in each column, multiplied by 2, give the numbers in the next column.
So, a factor of 144 can be found like this:

Multiply some 2’s and 3’s . The number of 2’s must be less than or equal to 4 (we can also choose to take no 2 at all). The number of 3’s must be less than or equal to 2 (or no 3 at all). Such factors, together with lgive all the factors.

For example, 24 is the product of three 2’s and one 3.
24 = 2 × 2 × 2 × 3
And 18 is the product of a single 2 and two 3’s.
Can you find the factors of 200 like this?
200 = 2 × 2 × 2 × 5 × 5
Make a table like this:
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 11
Answer:

Explanation:
factors of 200 = 2 × 2 × 2 × 5 × 5
Thus the numbers in each column, multiplied by 5, to give the numbers in the next column.

Find all the factors of the numbers below:

(i) 242
Answer:
1, 2 and 11 are the factors of 242
Explanation:
2 × 11 = 22
2 × (11 × 11) = 22 × 11 = 242

(ii) 225
Answer:
1, 3 and 5 are the factors of 225
Explanation:
5 × 5 = 25
(5 x 5) × (3 × 3) = 25 × 9 = 225

(iii) 400
Answer:
1, 2 and 5 are the factors of 400
Explanation:
5 × 5 = 25
2 x 2 x 2 x 2 = 16
(5 x 5) × (2 x 2 x 2 x 2) = 25 × 16 = 400

(iv) 1000
Answer:
1, 2 and 5 are the factors of 400
Explanation:
5 × 5 x 5 = 125
(5 x 5 x 5) × (2 x 2 x 2) = 125 × 8 = 400

We have found the factors of 144.
Now let’s try 144 × 5 = 720
720 = 2 × 2 × 2 × 2 × 3 × 3 × 5
We can separate the factors as those without 5 and those with 5.
The factors without 5 are factors of 144.
And these can be found as before.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 12
Multiplying all these by 5 gives the factors with 5.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 13

Let’s write all these factors of 720 in a single table:
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 14
What about 144 × 25 = 3600?
We can expand the factor table of 720 like this:
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 15

Factorize each of the numbers below as the product of primes and write all factors in a table.
Write also the number of factors of each.

(i) 72
Answer:
1, 2 and 3 are factors of 72

Explanation:
factorization of 72 is,
1 x 2 = 2
3 × 3 = 9
2 x 2 x 2  = 8
(3 x 3) × (2 x 2 x 2) = 9 × 8 = 72
So, 1, 2 and 3 are factors of 72.

(ii) 108
Answer:

Explanation:
factorization of 108 is,
2 × 2 = 4
(3 x 3 x 3) × (2 x 2) = 27 × 4 = 108
So, 1, 2, 3 are factors of 108

(iii) 300
Answer:

Explanation:
Factorization of 300 is,
3 x 2 x 2 x 5 x 5 x 1
= 3 x 4 x 25
= 300

(iv) 96
Answer:
1, 2 and  3 are factors of 96

Explanation:
factorization of 96 is 3 x 2 x 2 x 2 x 2 x 2 x 1.
factors of 96 are 1, 2 and 3.

(v) 160
Answer:

Explanation:
factorization of 160 is,
2 x 2 x 2 x 2 x 2 x 5 x 1 = 160

(vi) 486
Answer:
2 x 3 x 3 x 3 x 3 x 3 = 486
Explanation:
factorization of 486 is,
2 x 3 x 3 x 3 x 3 x 3 = 486
LCM of 486 is as follows,

(vii) 60
Answer:
1, 2 ,3 and 5 are the factors of 60

Explanation:
factorization of 60 is,
2 x 2 x 3 x 5 = 160
LCM of 60 is shown below,

(viii) 90
Answer:

Explanation:
factorization of 90 is,
2 x 3 x 3 x 5 = 90
factor of 90 are 1, 2, 3 and 5
LCM of 90 is shown below,

(ix) 150
Answer:

Explanation:
factorization of 150 is,
2 x 3 x 5 x 5 = 150
LCM of 150 is shown below,

(i) Find the number of factors of 6, 10, 15, 14, 21. Find some other numbers with exactly four factors.
Answer:
Factors of 6 are 1, 2, 3 and 6.
Factors of 10 are 1, 2, 5 and 10.
Factors of 15 are 1, 3, 5 and 15.
Factors of 14 are 1, 2, 7 and 14.
Factors of 21 are 1, 3, 7 and 21.
Explanation:
A factor is a number which divides the number without remainder.
Factors of 6 are 1, 2, 3 and 6.
1 x 6 = 6
2 x 3 = 6
3 x 2 = 6
6 x 1 = 6
Factors of 10 are 1, 2, 5 and 10.
1 x 10 = 10
2 x 5 = 10
5 x 2 = 10
10 x 1 = 10
Factors of 15 are 1, 3, 5 and 15.
1 x 15 = 15
3 x 5 = 15
5 x 3 = 15
15 x 1 = 15
Factors of 14 are 1, 2, 7 and 14.
1 x 14 = 14
2 x 7 = 14
7 x 2 = 14
14 x 1 = 14
Factors of 21 are 1, 3, 7 and 21.
1 x 21 = 21
3 x 7 = 21
7 x 3 = 21
21 x 1 = 21

(ii) Is it correct to say that any number with exactly four factors is a product of two distinct primes?
Answer:
No, this is not correct.
Explanation:
If any prime number then its 3rd power has exactly 4 divisors,
and is obviously not the product of two distinct primes.
Take 36, it has two prime factors, 2 and 3.

Number of factors Textbook Page No. 104

We know how to find all the factors of 64.
Without writing down all the factors, can we just find the number of factors?
64 = 2 × 2 × 2 × 2 × 2 × 2
We can take one 2, two 2’s, three 2’s and so on to get factors. How many such factors are there?
Here there are six 2’s. So we can take one to six 2’s, and 1 is also a factor.
6 + 1 = 7 factors in all.
Can we find the number of factors of243 like this?
243 = 3 × 3 × 3 × 3 × 3
How many 3’s?
Taking one 3, two 3’s, three 3’s and so on, how many factors do we get?
Together with 1?
5 + 1 = 6 factors in all.
If a number can be split as the repeated product of a single prime, how do we find the number of factors of that number quickly?
What if we have two primes?
For example, let’s take 64 × 3 = 192
192 = (2 × 2 × 2 × 2 × 2 × 2) × 3
1 and products of 2’s give 7 factors as above; these factors multiplied by 3 give another 7 factors. Altogether 14 factors.
How about one more 3?
So how many factors does 192 × 3 = 576 have?
576 = (2 × 2 × 2 × 2 × 2 × 2) × (3 × 3)

We can separate the factors of 576 like this.

(i) Factors without 3
1 2 4 8 16 32 64

(ii) Product of these by 3
3 6 12 24 48 96 192

(iii) Product of the first factors by two 3’s
9 18 36 72 144 288 576

7 of each type. 7 × 3 = 21 in all.
We can put this in a different way. Take the products of 2’s and 3’s separately.
576 = 64 × 9

Look at the three types of factors of 576 again

(i) 1, 2, 4, 8, 16, 32, 64 – Factors of 64
(ii) 3, 6, 12, 24, 48, 96, 192 – Products of the factors of 64 by the factor 3 of 9
(iii) 9, 18, 36, 72, 144, 288, 576 – Products of the factors of 64 by the factor 9 of 9

We can also say that the factors we write first are the product of the factors of 64 by the factor 1 of 9.

Thus the factors of 576 are the product of each factor of 64 by each factor of 9.
64 has 7 factors and 9 has 3 factors. So 64 × 9 = 576 has 3 groups of 7 factors.
That is 7 × 3 = 21 factors.
Like this, can we find how many factors 1000 has?
1000 = (2 × 2 × 2) × (5 × 5 × 5)

In this, 2 × 2 × 2 = 8 has 4 factors; and 5 × 5 × 5 = 125 also has 4 factors.
We can multiply each of the 4 factors of 8 by each of the 4 factors of 125 to get all factors of 1000. That is 4 groups of 4 factors, making 4 × 4 = 16 in all.
Now let’s see how many factors 3600 has.
3600 = (2 × 2 × 2 × 2) × (3 × 3) × (5 × 5)
2 × 2 × 2 × 2 = 16 has 5 factors, 3 × 3 = 9 and 5 × 5 = 25 have 3 factors each.

Multiplying each factor of 16 by each factor of 9 gives 5 × 3 = 15 factors of 16 × 9.
Multiplying each of these by factors of 25 give all factors of 16 × 9 × 25 = 3600.
That means 15 × 3 = 45 factors
(Look once more the factor table of 3600 done earlier).
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 17

The number 4 has 3 factors and number 6 has 4 factors. Can we say that 4 × 6 = 24 has 3 × 4 = 12 factors’? Multiply each factor of 4 by each factor of 6. Why did we get the number of factors wrong?
Answer:
Yes, 4 × 6 = 24 has 3 × 4 = 12 factors’.
Explanation:
factors of 4 = 1, 2, 4
factors of 6 = 1, 2, 3, 6
Let 48 be the number,
factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
4 is one of the factor of 48,
6 is one of the factor of 48.
4 x 6 = 24 is also a factor of 48.
let x be the number,
if 4 and 6 are 2 of the factors of x,
then 4 x 6 = 24 is also one of the factor of x.

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Textbook Page No. 107

Question 1.
The factor table of a number is given below. Some of the factors are written.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 19
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 18
Answer:

(i) What is the number with this factor table?
Answer:
1, 2 , 14, 100, 245, 490, 4900
Explanation:
1 x 1 = 1
1 x 2 = 2
7 x 2 = 14
7 x 7 x 5 = 245
2 x 2 x 5 x 5 x 7 x 7 = 4900

(ii) Fill in the numbers in the circles
Answer:
1, 14, 245, 4900
Explanation:
1 x 1 = 1
7 x 2 = 14
7 x 7 x 5 = 245
2 x 2 x 5 x 5 x 7 x 7 = 4900

(iii) Write the numbers below in the correct cells
4, 25, 140, 200
Answer:

Explanation:
Take the LCM of given numbers 4, 25, 140 and 200
or find the factors of all the numbers above to fit in the cells as shown below.

(iv) Which of the numbers below cannot be in the table?
32, 40, 50, 200, 300, 350
Answer:
32 and 300
can not be in the table
Explanation:
Due to non availability of factor Five 2’s and 3 in the table 32 and 300 can not be in the table

Question 2.
Find the number of factors of each of these numbers.
(i) 500
Answer:
12
Explanation:
find the LCM of 500
500 = 2x 53.
= (2 + 1) x (3 + 1)
= 3 x 4
= 12

(ii) 600
Answer:
24
Explanation:
find the LCM of 600
600 = 23 x 31 x 52.
= (3 + 1) x (1 + 1) x (1 + 2)
= 4 x 2 x 3
= 24

(iii) 700
Answer:
12
Explanation:
first take the LCM of 700
700 = 22 x 52 x 71.
= (2 + 1) x (1 + 2) x (1 + 1)
= 3 x 3 x 2
= 12

(iv) 800
Answer:
18
Explanation:
find the LCM of 800
800 = 25 x 52.
= (5 + 1) x (1 + 2)
= 6 x 3
= 18

(v) 900
Answer:
27
Explanation:
Find the LCM of 900
900 = 22 x 32 x 52.
= (2 + 1) x (2 + 1) x (2 + 1)
= 3 x 3 x 3
= 27

Question 3.
How many factors does a product of three distinct primes have? What about a product of 4 distinct primes?
Answer:
Product of 3 distinct primes have 8 factors.
Product of 4 distinct primes have 12 factors.
Explanation:
To find the factors of product of three distinct primes have and four distinct primes have.
The factors of product of three distinct primes have four factors that include 1.
Let us consider a product of the primes 2, 3 and 5.
2 × 3 × 5 = 30
Factors of 30 are 1 , 2 , 3, 5 , 6 , 10 , 30 where the prime factors are 2, 3 and 5.
Hence there are 8 factors in total for a number that is the product of three distinct primes.
Let us consider a product of the primes 2, 3, 5 and 7.
2 × 3 × 5 x 7 = 210
The factors of product of four distinct primes have five factors that include 1.
The factors of 210 are 1, 2, 3, 5, 6, 7, 15, 30, 42, 70, 105 and 210.
So, it has twelve factors.
Hence there are 8 factors in total for a number that is the product of three distinct primes.

Question 4.
i) Find two numbers with exactly five factors.
Answer:
16 and 81
Explanation:
A factor is a number which divides the number without remainder.
factors of 16 = 1, 2, 4, 8 and 16.
factors of 81 = 1, 3, 9, 27, 81.

ii) What is the smallest number with exactly five factors?
Answer:
16
Explanation:
A factor is a number which divides the number without remainder.
factors of 16 = 1, 2, 4, 8 and 16.

Question 5.
How many even factors does 3600 have?
Answer:
36 factors.
Explanation:
Number of even factors = no. of total factors – no. of odd factors.
Prime Factorization of 3600 =  24 x 32 x 52
Total factors of 3600 = 45;
Total number of odd factors of 3600 is (2 + 1)(2 + 1) = 3 × 3 = 9
Even factors = 45 – 9 = 36

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