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# Really need some help!

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Apr 7, 2020
edited by somehelpplease  May 8, 2020

#1
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If you divide 1/(5^n) out, you'll find that the digits right of the decimal point are equal to n. Based on that, the smallest positive integer n such that f(n)>10 is 11.

Hope it helps!

Apr 7, 2020
#2
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I'm sorry, but it said that was incorrect. You said the answer was 11?

#3
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I thought it was, but please let me see if I can fix this answer.

Edit: I realize that I did the number of digits and not the sum of the digits, please let me fix this. Thanks!

HELPMEEEEEEEEEEEEE  Apr 7, 2020
edited by HELPMEEEEEEEEEEEEE  Apr 7, 2020
#4
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No worries. :D

#5
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Ok, new answer... 1/(5^2) has a pattern where the digits to the right of the decimal place are powers of 2. 2, 4, 8, 16, 32... and so on. Finding the sum of these digits we get 2=2, 4=4, 8=8, 1+6=7, 3+2=5, 6+4=10, and 1+2+8=11. 128 is 2^7 so your answer is 7.

Hope it helps!

Apr 7, 2020
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Yup, it was correct. Thanks for re-working it out and also the explanation. :D Really appreciate it, and have a good day!