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let the sum of the first five numbers be X. so the last digit can be filled in 10 ways. let it be Y.
assume X is divisible by 5. for X+Y to be divisible, only 0, or 5 can be the last number.
similarly if X not divisible by 5, X+Y will be divisible by only two nos out of 10 different last digit.
hence number of 6 digit numbers in which the sum of the digits is divisible by 5, is simply 1/5th the no of 6 digit nos.
which is 9*10^5 /5 = 180000.
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Moderation Team
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