NCERT Solutions Class 6th Maths Chapter – 3 Playing With Numbers All Examples

Example 1. Write all the factors of 68.

Solution –

All factors of 68

68 = 1 × 68
68 = 2 × 34
68 = 4 × 17
68 = 17 × 4

Stop here, because 4 and 17 have occurred earlier.
Thus, all the factors of 68 are 1, 2, 4, 17, 34 and 68.

Example 2. Find the factors of 36.

Solution –

All factors of 36

36 = 1 × 36
36 = 2 × 18
36 = 3 × 12
36 = 4 × 9
36 = 6 × 6

Stop here, because both the factors (6) are same. Thus, the factors are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Example 3. Write first five multiples of 6.

Solution –

The required multiples are- 6×1= 6, 6×2 = 12, 6×3 = 18, 6×4 = 24, 6×5 = 30 i.e. 6, 12, 18, 24 and 30.

Example 4. Write all the prime numbers less than 15.

Solution –

By observing the Sieve Method, we can easily write the required prime numbers as 2, 3, 5, 7, 11 and 13.

Example 5. Find the common factors of 75, 60 and 210.

Solution –

Factors of 75 are 1, 3, 5, 15, 25 and 75.

Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 30 and 60.

Factors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.

Thus, common factors of 75, 60 and 210 are 1, 3, 5 and 15.

Example 6. Find the common multiples of 3, 4 and 9.

Solution –

Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, ….

Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48,…

Multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, 81, …

Clearly, common multiples of 3, 4 and 9 are 36, 72, 108,…

Example 7. Find the prime factorisation of 980.

Solution –

We proceed as follows – We divide the number 980 by 2, 3, 5, 7 etc. in this order repeatedly so long as the quotient is divisible by that number.Thus, the prime factorisation of 980 is 2 × 2 × 5 × 7 × 7.

Example 8. Find the LCM of 12 and 18.

Solution –

We know that common multiples of 12 and 18 are 36, 72, 108 etc. The lowest of these is 36. Let us see another method to find LCM of two numbers.

The prime factorisations of 12 and 18 are-
12 = 2 × 2 × 3; 18 = 2 × 3 × 3

In these prime factorisations, the maximum number of times the prime factor 2 occurs is two; this happens for 12. Similarly, the maximum number of times the factor 3 occurs is two; this happens for 18. The LCM of the two numbers is the product of the prime factors counted the maximum number of times they occur in any of the numbers. Thus, in this case LCM = 2 × 2 × 3 × 3 = 36.

Example 9. Find the LCM of 24 and 90.

Solution –

The prime factorisations of 24 and 90 are-
24 = 2 × 2 × 2 × 3; 90 = 2 × 3 × 3 × 5

In these prime factorisations the maximum number of times the prime factor 2 occurs is three; this happens for 24. Similarly, the maximum number of times the prime factor 3 occurs is two; this happens for 90. The prime factor 5 occurs only once in 90.

Thus, LCM = (2 × 2 × 2) × (3 × 3) × 5 = 360.

Example 10. Find the LCM of 40, 48 and 45.

Solution –

The prime factorisations of 40, 48 and 45 are-
40 = 2 × 2 × 2 × 5
48 = 2 × 2 × 2 × 2 × 3
45 = 3 × 3 × 5

The prime factor 2 appears maximum number of four times in the prime factorisation of 48, the prime factor 3 occurs maximum number of two times in the prime factorisation of 45, The prime factor 5 appears one time in the prime factorisations of 40 and 45, we take it only once.

Therefore, required LCM = (2 × 2 × 2 × 2)×(3 × 3) × 5 = 720
LCM can also be found in the following way.

Example 11. Find the LCM of 20, 25 and 30.

Solution –

We write the numbers as follows in a row-

So, LCM = 2 × 2 × 3 × 5 × 5.

(A) Divide by the least prime number which divides atleast one of the given numbers. Here, it is 2. The numbers like 25 are not divisible by 2 so they are written as such in the next row.

(B) Again divide by 2. Continue this till we have no multiples of 2.

(C) Divide by next prime number which is 3.

(D) Divide by next prime number which is 5.

(E) Again divide by 5.

Example 12. Two tankers contain 850 litres and 680 litres of kerosene oil respectively. Find the maximum capacity of a container which can measure the kerosene oil of both the tankers when used an exact number of times.

Solution –

The required container has to measure both the tankers in a way that the count is an exact number of times. So its capacity must be an exact divisor of the capacities of both the tankers. Moreover, this capacity should be maximum. Thus, the maximum capacity of such a container will be the HCF of 850 and 680.

It is found as follows-

The common factors of 850 and 680 are 2, 5 and 17.

Thus, the HCF of 850 and 680 is 2 × 5 × 17 = 170.

Therefore, maximum capacity of the required container is 170 litres.

It will fill the first container in 5 and the second in 4 refills.

Example 13. In a morning walk, three persons step off together. Their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that all can cover the same distance in complete steps?

Solution –

The distance covered by each one of them is required to be the same as well as minimum. The required minimum distance each should walkwould be the lowest common multiple of the measures of their steps. Can you describe why? Thus, we find the LCM of 80, 85 and 90. The LCM of 80, 85 and 90 is 12240.

The required minimum distance is 12240 cm.

Example 14. Find the least number which when divided by 12, 16, 24 and 36 leaves a remainder 7 in each case.

Solution –

We first find the LCM of 12, 16, 24 and 36 as follows-

Thus, LCM = 2 × 2 × 2 × 2 × 3 × 3 = 144

144 is the least number which when divided by the given numbers will leave remainder 0 in each case. But we need the least number that leaves remainder 7 in each case.

Therefore, the required number is 7 more than 144. The required least number = 144 + 7 = 151.