Question
What is the largest sum of m and n such that
β4m932n6β is divisible by 12?Solution
ATQ,
Divisible by 12 β divisible by 3 and 4
Last two digits: βn6β divisible by 4 β 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 β m = 1, 4, 7
Max m = 7
β max (m + n) = 7 + 8 = 15
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