If 21358AB is divisible by 99, then it is divisible by 9 and 11.
If 21358AB is divisible by 9 then the sum of the digits is divisible by 9.
2 + 1 + 3 + 5 + 8 + A + B = 9 × 3
say,
⇒ 19 + A + B = 27
⇒ A + B = 27 - 19 = 8
A + B = 8 …… (1)
If 21358AB is divisible by ‘11’ then the difference of sum of even and odd digits will be divisible by’11’.
2 1 3 5 8 A B
∴ (2 + 3 + 8 + B) – (1 + 5 + A) = 11 × 1
say,
⇒ 13 + B – 6 – A = 11
⇒ B – A = 11 – 7 = 4 ……. (2)
From (1) & (2) A = 2, B = 6
∴ The required number is 21358AB = 2135826 which is divisible by 99.
