A 4-digit number x92x is divisible by 36 and a 7-digit number y569z04 is divisible by 11. For the maximum value of (y + z), what is the value of (2x + y +z)?
Show answer & explanation
Correct answer: A
Explanation
The correct answer is option 1 i.e. 31.
4-digit number x92x divisible by 36 (Divisible by both 4 and 9)
2x must be divisible by 4. So, the possible values for x: 0, 4, 8
Sum of digits = x + 9 + 2 + x = 2x + 11
If x = 8 then 2x + 11 = 27 → Divisible by 9
Valid number: 8928
7-digit number y569z04 divisible by 11 (maximize y + z)
(y + 6 + z + 4) - (5 + 9 + 0) = (y + z + 10) - 14 = y + z - 4
If y + z = 15 then y + z - 4 = 11 → Divisible by 11
Since y is the first digit, it must be as large as possible.
Let y = 9, then z = 6
Valid number: 9569604
Now, (2x + y + z) = 2(8) + 9 + 6 = 16 + 15 = 31