Question
If '38k721' is divisible by 3, then what is the highest
possible 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 '38k721' = 3 + 8 + k + 7 + 2 + 1 = (k + 21)
(k + 21) is divisible by 3 when k = 0, 3, 6 or 9.
Therefore, maximum value of 'k' = 9
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