Question
Find the difference between minimum and maximum value
of 'i' such that '8i2470' 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 '8i2470' = (8 + i + 2 + 4 + 7 + 0) = i + 21.
So, i + 21 should be divisible by 3.
Possible values of 'i' = 0, 3, 6, 9
Minimum value of 'i' = 0
Maximum value of 'i' = 9
Required difference = 9 – 0 = 9
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