Question
Find the digit sum of the largest three-digit number that leaves 1 as a remainder when divided by 8, 6, and 10.
Solution
ATQ,
LCM of (8, 6, and 10) = 120
Largest 3-digit number divisible by 120 = 120 × 8 = 960
So, N = 960 + 1 = 961
Sum of digits of 'N' = 9 + 6 + 1 = 16
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