AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.2 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.2

Question 1.

If 345 A 7 is divisible by 3,supply the missing digit in place of ‘A’.

Solution:

If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.

∴ 345A7 ⇒ 3 + 4 + 5 + A + 7 = 19 + A

19 + A = 3 x 7

⇒ A = 21 – 19 = 2 ⇒ A = 24 – 19 = 5

A + 19 = 3 x 8

⇒ A = 24 – 19 = 5

A + 19 = 3 x 9

⇒ A = 27 – 19 = 8

∴ A = {2,5,8}

Question 2.

If 2791 A,is divisible by 9, supply the missing digit in place of ‘A’.

Solution:

If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.

∴ 2791A = 2 + 7 + 9 + 1 + A = 9 x 3

⇒ 19 + A = 9 x 3 = 27

⇒ A = 27 – 19 = 8

∴ A = 8

Question 3.

Write some numbers which are divisible by 2,3,5,9 and 10 also.

Solution:

90, 180, 270. are divisible by 2, 3, 5, 9 and 10.

[∵ The L.C.M. of 2, 3, 5, 9, 10 is 90]

Question 4.

2A8 is a number divisible by 2, what might be the value of A’?

Solution:

If the units digit of a number be 0, 2, 4, 6, 8 then it is divisible by 2.

∴ 2A8 is divisible by 2 for any value of A.

∴ A = (0, 1, 2 ………………….9)

Question 5.

50B is a number divisible by 5, what might be the value of B?

Solution:

Given number is 50B.

The units digit of a number ¡s either ‘0’ or 5, then it is divisible by 5.

∴ 500 → \(\frac { 0 }{ 5 }\) (R = 0)

505 → \(\frac { 5 }{ 5 }\) (R = 0)

∴ B = {0, 5}

Question 6.

2P is a number which is divisible by 2 and 3, what is the value of P

Solution:

The given number is 2P.

If 2P is divisible by 2, 3 then 2P should be a multiple of 6. [ ∵ L.C.M. of 2, 3 is 6]

∴ 2P = 24, 30 ………….

24 → 2 + 4 → \(\frac { 6 }{ 3 }\) (R = 0)

∴ P = 4

Question 7.

54Z leaves remainder 2 when divided by 5 , and leaves remainder 1 when divided by 3, what is the value of Z’?

Solution:

If 54Z is divisible by 3 then the sum of the digits of the number is divisible by 3.

According to problem 54Z is divisible by 3 and leaves remainder 1’.

∴ 5 + 4 + Z = (3 x 4) + 1

= 9 + Z = 13

∴ Z = 4(or)

9 + Z = (3 x 5) + 1

9 + Z = 16

Z = 7

If 54Z is divisible by 5 then Z should be equal to either ‘0’ or ‘5’.

∴ 54(0 + 2) = 542 (Z = 2)

54(0 + 7) = 547 (Z = 7)

∴ From the above two cases

Z = 7

∵ 547 → \(\frac{7}{5}\)(R = 2)

Question 8.

27Q leaves remainder 3 when divided by 5 and leaves remainder 1 when divided by 2, what is the remainder when it is divided by 3?

Solution:

27Q is divided by 5 gives the remainder 3

Le.,27Q = 27 (0 + 3) = 273(Z = 3)(T)

= 27 (0 + 8) = 278 (Z = 8)

27Q is divided by 2 gives the remainder 1.

i.e., 27Q = 27(0 + 1) = 271 (Z = 1)

27Q = 27 (0 + 3) = 273 (Z = 3) (T)

∴ From above situations Z = 3

∴ 27Q = 273→ 2 + 7 + 3 → \(\frac{12}{3}\)(R = 0)

∴ 273 is divisible by 3 and gives the remainder 0’.