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Let $a$ be a positive integer. If $S(a)$ is the sum of 3-digit part of $a$, then prove $a$ is divisible by 37, if and only if $S(a)$ is divisible by 37.

I mention a, is the sum of 3 digit parts of a. For example, $a=5987654321$, then $S(a)=5 + 987 + 654 + 321 =1967$. I think this is very similar to the divisibility by 3, like all the digits sum is divisible by 3 then the number is divisible by 3. I think it can use decimal representation and then prove like that. And use congruences to help proove it. Can someone help judge if this is right?

Carlos
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Kölle
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    Note: $999=37\times27$, so $1000\equiv1\bmod37$ (just like $10\equiv1\bmod3$) – J. W. Tanner Nov 19 '20 at 23:12
  • right. the rule of three works because if $N=\sum a_k10^k$ and $10\equiv 1 \pmod 3$.....(so $N\equiv \sum a_k\pmod 3$). So if $b_k$ are the "3 part digits" then $N = \sum b_k (1000)^k$ (that's the first maybe not obvious thing to realize) and as it turns out..... $1000 \equiv 1 \pmod 37$ (that's the second not obvious, but easy to verify, thing to realize) .... – fleablood Nov 20 '20 at 00:12
  • Yes, since $, 37\mid 10^3-1,$ it's a radix $10^3$ analog of casting out nines in radix $10,,$ cf. this answer in the dupe. – Bill Dubuque Nov 21 '20 at 23:49

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You're right.

$a_0+a_1\times10+a_2\times10^2+a_3\times10^3+\cdots+a_n\times10^n$

$\equiv a_0+a_1+a_2+a_3+\cdots+a_n\bmod3,$ because $1\equiv10\equiv10^2\equiv10^3\equiv\cdots\equiv10^n\bmod 3$,

and $b_0+b_1\times1000+b_2\times1000^2+b_3\times1000^3+\cdots+b_n\times1000^n$

$\equiv b_0+b_1+b_2+b_3+\cdots+b_n\bmod37$,

because $1\equiv1000\equiv1000^2\equiv1000^3\equiv\cdots\equiv1000^n\bmod37$.

($1000=999+1=27\times37+1\equiv1\bmod 37$.)

J. W. Tanner
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