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Trailing Zeros Quest (Posted on 2025-01-05)

Let n denote the smallest positive integer such that 11n-3 ends in at least 1000 zeros. Compute the sum of the digits of n.

No Solution Yet Submitted by Danish Ahmed Khan

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Solution

| Comment 1 of 2

The smallest number would end in 3 preceded by 999 zeros.

The smallest digit before this would be 8 since 8+3=11

8/11 = 0.727272...

So 8[999 zeros]3 when divided by 11 is

n=727272...73

with 500 7's, 499 2's and a 3 at the end.

The answer is 9*500+1 = 4501

Posted by Jer on 2025-01-05 18:14:49

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