Question
If the number '6k3942' is divisible by 3, then find the
maximum value of 'k'.Solution
ATQ,
A number is divisible by 3 when the sum of its digits is divisible by 3.
Sum of digits of '6k3942' = 6 + k + 3 + 9 + 4 + 2 = (k + 24)
(k + 24) is divisible by 3 when k = 0, 3, 6 or 9
Therefore, maximum value of 'k' = 9
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