# ML Aggarwal CBSE Solutions Class 6 Math Third Chapter Playing with Numbers Exercise 3.2

### ML Aggarwal CBSE Solutions Class 6 Math Third Chapter Playing with Numbers Exercise 3.2

(1) Which of the following numbers are divisible by 5 or by 10:

Solution: The number having last digit 0 or 5 divisible by 5. & The number having last digit 0 divisible by 10.

Here – (i) 3725 – Divisible 5.

(ii) 48970 – Divisible by 5 and 10 both

(iii) 56823 – Not divisible by 5 or 10

(iv) 760035 – Divisible by 5

(v) 7893217 – Not divisible by 5 or 10

(vi) 4500010 – Divisible by 5 / 10 both

(2) Which of the following numbers are divisible by 2, 4 or 8

Solution: *The numbers ending with 0, 2, 4, 6, 8 then the number divisible by 2.

*A number divisible by 4 if the number formed by its last two digits is divisible by 4

*A number divisible by 8 if the number formed by its last three digits is divisible by 8.

Divisible by 2 : (i), (ii), (iii), (iv), (vi)

Divisible by 4: (ii), (iii), (iv), (vi)

Divisible by 8: (ii), (iii), (vi)

(3) Which of the following numbers are divisible by 3 or 9:

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

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

Divisible by 3: (i), (ii), (iii), (iv)

Divisible by  9: (ii), (iii), (vi)

(4) Examine the following numbers for divisibility by 11:

Solution: A number is divisible by 11 if the sum of numbers in blocks of two digits from right to left is divisible by 11

(i) 10428 = 28 + 04 + 1 = 33, which is divisible by 11

∴ The given number 10428 is divisible by 11.

(ii) 70169803

03 + 98 + 16 + 70 = 187, which is divisible by 11

∴ The given number 70169803 is divisible by 11

(iii) 7136985

85 + 69 + 13 + 7 = 174, which is not divisible by 11

∴ The given number 7136985 is not divided by 11

(5) Examine the following numbers for divisibility by 6:

Solution: A number divisible by 2 and 3 therefore the number divisible by 6

(i) 93573

Given number is not divisible by 6 as because, it is not divided by 2.

(ii) 217944

Since the unit’s digit in the given number is 4, so the given number is divisible by 2.

Also the sum of the digits  = 2 + 1 + 7 + 9 + 4 + 4 = 27, which is divisible by 3, so the given number is divisible by 3.

Therefore, the given number 217944 is divisible by 6

(iii) 5034126

Since the unit digit in the given number  is 6, so the given number is divisible by 2.

Also the sum of the digits = 5 + 0 + 3 + 4 + 1 + 2 + 6 = 18, which is divisible by 3

Therefore, the given number  5034126 is divided by 6.

(iv) 901352

The sum of the digits – 9 + 0 + 1 + 3 + 5 +2 = 20, Not divided by 3

Therefore, the given number is not divided by 6

(v) 639210

Since the unit digit in the given number is 0, which is divided by 2.

The sum of the digits – 6 + 3 + 9 + 2 + 1 + 0 = 21, divided by 3

Therefore, the given number is divided by 6.

(vi) 1790184

Since the unit digit of the given number is 4, therefore the number is divided by 2

The sum of the digit = 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30, divided by 3

Therefore, the given number  is divisible by 6.

(6) In each of the following replace * by a digit so that the number formed is divisible by 9:

(i) 4710*82

Sum of the given digit = 4 + 7 + 1 + 0 + 8 + 2 = 22

If we add 5, it becomes 27, which is divisible by 9

Therefore, * is to be replaced by 5

The number formed is 4710582, which is divisible by 9

(ii)70*356722

The sum of the digits 7 + 0 + 3 + 5 + 6 + 7 + 2 + 2 = 32

If we add 4 with the sum, therefore the number become 36, which is divided by 9

Therefore * is to be replaced by 4

The number is -704356722

(7) In each of the following replace * by (i) the smallest digit (ii) the greatest digit so that the number formed is divisible by 3:

Solution: (a) 4*672

Sum of digits (except *) = 2 + 7 + 6 + 4 = 19,

If * is replaced by 2, then

Sum of the digits 19 + 2 = 21, which is de3vided by 3

So, * is to be replaced by 2

(b) 4756*2

Sum of digits (except *) = 4 + 7 + 5 + 6 + 2 = 24

So, we can replaced * by O

(8) In each of the following ‘*’ by a digit so that the number formed is divisible by 11:

(i) 8*9484

Solution: In 8*9484, * occurs at odd places.

Sum of digits at odd places (except *) = 4 + 4 = 8

Sum of digits at even places = 8 + 9 + 8 = 25

Their difference = 25 – 8 = 17

If * is replaced by 6, then sum of digits at odd places = 4 + 4 + 6 = 16

Then their difference = 25 – 16 = 11, which is divisible by 11 so that the number formed will be divisible by 11

∴ * is to be replaced by the digit 6

(ii) 9*53762

Solution: * occurs at even places.

Sum of digits at odd places = 2 + 7 + 5 + 9 = 23

Sum of digits at even places (Except *) = 6 + 3 = 9

Their difference = 23 – 9 = 14

If * is replaced by 3

Then sum of digits at even places = 6 + 3 + 3 = 12

Then their difference = 23 – 12 = 11, which is divisible by 11

∴ * is to be replaced by 3

(10) Which of the following numbers are prime:

(i) 101

Solution: (i) As 10 x 10 = 100 < 101 and 11 x 11 = 121 > 101

So 10 is the largest number  such that 10 x 10 ≤ 101. The prime numbers less than or equal to 10 are 2, 3, 5, 7

Note that, 101 is not divisible by any of 2, 3, 5, 7

Therefore, 101 is a prime number.

(ii) 251

Solution: As 15 x 15 = 225 < 251 and 16 x 16 = 256 > 251

So 15 is the largest number such that 15 x 15 ≤ 251. The prime numbers less than or equal to 15 is 2, 3, 5, 7, 11, 13

Here this number 251 is not divisible by 2, 3, 5, 7, 11, 13

Therefore, 251 is a prime number.

(iii) 323

Solution: As 17 x 17 = 289 > 323 and 18 x 18 = 324 > 323

So 17 is the largest number such that 17 x 17 ≤ 323. The prime numbers less than or equal to 17 is 2, 3, 5, 7, 11, 13

Here the number 323 is not divided by 2, 3, 5, 7, 11, 13

Therefore, 323 is a prime number

(iv) 397

Solution: As 19 x 19 = 361 < 397 and 20 x 20 = 400 > 397

So, 19 is the largest number such that 19 x 19 ≤ 397. The prime numbers less than or equal to 19 is 2, 3, 5, 7 , 11, 17, 19

Here the number 397 is not divided by 2, 3, 5, 7 , 11, 17, 19

So, 397 is a prime number.

(11) Determine 372645 is divisible 45.

Solution: Here, 45  = 5 x 9

The number 372645 is divided by both 5 and 9.

So, 372645 is divisible 45.

(12) A number is divisible by 12. By what other numbers will that be divisible?

Solution: The other number is = 1, 2, 3, 4, 6

(13) A number is divisible by 3 and 8 both. By which other numbers will that number be always divisible?

Solution: Let a natural number, say n, be divisible by both 3 and 8.

As 3 and 8 are co prime numbers, using property 2, n is divisible by 3 x 8 = 24 i.e. 24.

Thus. The given number is always divisible by 24

In fact, this given number must be divisible by all the factors of 24 (using property 1).

The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24

(14) State whether the following statements are true (T) or false (F):

(i) False

(ii) False

(iii) True

(iv) True

(v) True

(vi) False

(vii)True

(viii) False

Updated: September 30, 2019 — 12:10 pm