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Prove that the repeated sum of the digits of a perfect square can only be 1,4,7 or 9. Example- 169- Sum-16, Sum of digits of 16 = 7. Please try to explain without log, mods. I am a grade 10th student. Thank You.

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marked as duplicate by Dietrich Burde, user91500, I am Back, Lord Shark the Unknown, Dando18 Aug 20 '17 at 23:22

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  • $\begingroup$ Yes, the question is incorrect. It is correct if you do repeated digit sums - $625\to 13\to 4$. $\endgroup$ – Thomas Andrews Aug 20 '17 at 18:21
  • $\begingroup$ Another answer here $\endgroup$ – Joffan Aug 21 '17 at 1:03
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Lemma 1: a natural number and the sum of its digits yield the same remainder when you divide them by $9$.

Lemma 2: if $n$ is a perfect square, the remainder of $n/9$ is $0,1,4$ or $7$.

Lemma 3: the repeated sum of the digits of a perfect square is $1,4,7$ or $9$.

It is not very hard to prove the first two lemmas, and the third should come easily with the first two.

Hint for Lemma 1: every natural number $m$ is $(a_k10^k-a_{k-1}10^{k-1}+\cdots+10a_1+a_0)$, where the $a_j$'s are its digits. Is $m-(a_k+a_{k-1}+\cdots+a_1+a_0)$ a multiple of $9$?

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