Question
Find the difference between minimum and maximum value
of 'f' such that '2f4896' 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 '2f4896' = (2 + f + 4 + 8 + 9 + 6) = f + 29.
So, f + 29 should be divisible by 3.
Possible values of 'f' = 2, 5, 8
Minimum value of 'f' = 2
Maximum value of 'f' = 8
Required difference = 8 – 2 = 6
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