​If a 9-digit number $$9834\space x\space 97\space y\space 4$$​ is divisible by 88, then what is the maximum possible value of $$(3x+2y)$$​?​

Correct option is A

Given:

9-digit number : 9834 x 97 y 4

Divisor : 88.

And we are asked to find the maximum possible value of (3x + 2y).

Solution:

A number is divisible by 8 if its last 3 digits form a number divisible by 8.
So, for number: 9834x97y4,
last 3 digits = 7y4
We now find the values of y such that 7y4 is divisible by 8:
​​​Test these:
704 ÷ 8 = 88 
744 ÷ 8 = 93 
784 ÷ 8 = 98 
So, valid y values: 0, 4, 8

Divisibility by 11
Digits: 9 8 3 4 x 9 7 y 4
Odd positions: 1, 3, 5, 7, 9 → 9, 3, x, 7, 4 → sum = 23 + x
Even positions: 2, 4, 6, 8 → 8, 4, 9, y → sum = 21 + y

Condition for divisibility by 11:
(23 + x) - (21 + y) ≡ 0 mod 11
→ x - y + 2 ≡ 0 mod 11
→ x - y ≡ -2 ≡ 9 mod 11
So, x - y = 9

Try values of y = 0, 4, 8:

y = 0 → x ≡ 9 mod 11 → x = 9 → 3x + 2y = 3(9) + 2(0) = 27

y = 4 → x ≡ 13 ≡ 2 mod 11 → x = 2 → 3x + 2y = 3(2) + 2(4) = 6 + 8 = 14

y = 8 → x ≡ 17 ≡ 6 mod 11 → x = 6 → 3x + 2y = 3(6) + 2(8) = 18 + 16 = 34

Final Step: Choose maximum of (3x + 2y)
y = 0 → 27

y = 4 → 14

y = 8 → 34

Correct Answer: (a) 34