For a number to be divisible by 36, it must have two 2s and two 3s i.e. to be divisible by 4 and 9. So an even number with the last two digits divisible by 4 and the sum of digits to be divisible by 9.
The sum of digits is: 5 + m + 1 + 5 + n = 11 + m + n (divisible by 9 and the last two digits (5,n) divisible by 4)
A number is divisible by 4, if the last two digits are divisible by 4. So, that means, the number has to end with 52 or 56.
5 + m + 1 + 5 + 2 = 13 + M
5 + m + 1 + 5 + 6 = 17 + N
a number is divisible by 9 if the sum of the digits is divisible by 9. However, the question asks for | m - n | to be maximum.
5 + m + 1 + 5 + 2 = 13 + M
The max value of | m - n | in this case is possible when m-9. But the number wouldn't be divisible by 9. The maximum possible, keeping into perspective the divisibility by 9, would be when m=5. In which case | m - n | would be 3.
5 + m + 1 + 5 + 6 = 17 + N
In the other case, similarly, the maximum value of | m - n | would be when m= 0. But then, the number wouldn't be divisible by 9. So we are looking for a number that is as far from 6 as possible. 1 suffices the condition.
Hence, the maximum possible value for |m-n| = 5
Answer is C.
Shoot me if I'm incorrect. I'm not confident though. Generally, I don't get answers right.
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