Question
Find the difference between minimum and maximum value of
'k' such that '6k8194' is always divisible by 3.Solution
A number is divisible by 3 when the sum of its digits is divisible by 3.
Sum of digits of '6k8194' = (6 + k + 8 + 1 + 9 + 4) = k + 28.
So, k + 28 should be divisible by 3.
Possible values of 'k' = 2, 5, 8
Minimum value of 'k' = 2
Maximum value of 'k' = 8
Required difference = 8 – 2 = 6
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