(i) The greatest 2 digit number 9A is divisible by 2. (ii) The least number 567A is divisible by 3.

Question

Find A as required

(i) The greatest 2 digit number 9A is divisible by 2.

(ii) The least number 567A is divisible by 3.

(iii) The greatest 3 digit number 9A6 is divisible by 6.

(iv) The number A08 is divisible by 4 and 9.

(v) The number 225A85 is divisible by 11.

Answer 1

(i) A number is divisible by 2 if it is an even number.

Greatest 2 digit even number is 98.

∴ A = 8

(ii) A number is divisible by 3 if the sum of its digits is divisible by 3

Sum of digits of 567A = 5 + 6 + 7+ A = 18 + A

∴ 18 is divisible by 3

∴ A may be 0

The number will be 5670

(iii) A number is divisible by 6 if it is divisible by both 2 and 3

9A6 is even and so divisible by 2

If A = 9 then the sum of digits will be = 24 which is divisible by 3.

The number will be 996 and A = 9

(iv) 08 is divisible by 4, so A08 is divisible by 4.

If A = 1 then the sum of digits will be 9 which is divisible by 9.

The number will be 108 and A = 1

(v) 5 + A + 2 – (8 + 5 + 2) 

= 7 + A – 15 

= -8 + A

∴ A = 8