Question
What is the largest value of 'k' that makes the number
'415k27' divisible by 3?Solution
ATQ,
A number is divisible by 3 when the sum of its digits is divisible by 3.
Sum of digits of '415k27' = 4 + 1 + 5 + k + 2 + 7 = (k + 19)
(k + 19) is divisible by 3 when k = 2, 5, 8
Therefore, maximum value of 'k' = 8
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