Question

(i) Find the value of the digit ‘a’ in 3a9 is divisible by 9
(ii) How many pairs of (a, b) are possible, when a 5 digit number 3a7b9 is divisible by 9.

Answer 1

(i) For the number to be divisible by 9 the sum of digits of the number must be a multiple of 9.
Sum of the digits of
$3a9=3+a+9=12+a$
(12 + a) should be a multiple of 9 to satisfy the condition. The nearest multiple of 9 after 12 is 18.
Therefore, 12 + a = 18
So, a = 18 – 12 = 6
(ii) For the number to be divisible by 9 the sum of digits of the number must be a multiple of 9.
Sum of the digits of
$3a7b9=3+a+7+b+9=19+a+b$.
(19 + a + b) should be a multiple of 9 to satisfy the condition. The nearest multiple of 9 after 19 are 27 and 36.
Therefore, 19 + a + b = 27 or 19 + a + b = 36.

So, a + b = 27 – 19 = 8 or
a + b = 36 – 19 = 17
So possible pairs of (a, b) are (8, 0), (7, 1), (6, 2), (5, 3), (4, 4), (3, 5), (2, 6), (1,7), (0, 8), (9, 8) and (8, 9).
Totally, 11 pairs of (a, b) are possible.