Question
What is the largest sum of m and n such that
‘4m932n6’ is divisible by 12?Solution
ATQ,
Divisible by 12 ⇒ divisible by 3 and 4
Last two digits: ‘n6’ divisible by 4 → n = 0, 2, 4, 6, 8
Take maximum n = 8
Sum = 4 + m + 9 + 3 + 2 + 8 + 6 = 32 + m
32 + m divisible by 3 → m = 1, 4, 7
Max m = 7
⇒ max (m + n) = 7 + 8 = 15
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