Question
Find the difference between minimum and maximum value
of 'g' such that '9g5621' 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 '9g5621' = (9 + g + 5 + 6 + 2 + 1) = g + 23.
So, g + 23 should be divisible by 3.
Possible values of 'g' = 2, 5, 8
Minimum value of 'g' = 2
Maximum value of 'g' = 8
Required difference = 8 – 2 = 6
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