Question
Find the difference between minimum and maximum value
of 'j' such that '7j5302' 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 '7j5302' = (7 + j + 5 + 3 + 0 + 2) = j + 17.
So, j + 17 should be divisible by 3.
Possible values of 'j' = 1, 4, 7
Minimum value of 'j' = 1
Maximum value of 'j' = 7
Required difference = 7 – 1 = 6
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