Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999

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Theorem

Let $n$ be an integer which has at least $3$ digits when expressed in decimal notation.

Let the digits of $n$ be divided into groups of $3$, counting from the right, and those groups added.


Then the result is equal to a multiple of $999$ if and only if $n$ is divisible by $999$.


Proof


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The mistake is either and conversely or equal to $999$, since $999 \, 999$ is an easy counterexample.

Here we will show that the result is equal to a multiple of $999$ if and only if $n$ is divisible by $999$.


Write $n = \ds \sum_{i \mathop = 0}^k a_i 10^{3 i}$, where $0 \le a_i < 1000$.

This divides the digits of $n$ into groups of $3$.

Then the statement is equivalent to:

$999 \divides n \iff 999 \divides \ds \sum_{i \mathop = 0}^k a_i$

This statement is true since:

\(\ds n\) \(=\) \(\ds \sum_{i \mathop = 0}^k a_i 10^{3 i}\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 0}^k a_i 1000^i\)
\(\ds \) \(\equiv\) \(\ds \sum_{i \mathop = 0}^k a_i 1^i\) \(\ds \pmod {999}\) Congruence of Powers
\(\ds \) \(\equiv\) \(\ds \sum_{i \mathop = 0}^k a_i\) \(\ds \pmod {999}\)

$\blacksquare$


Examples

\(\ds 4 \times 999\) \(=\) \(\ds 3996\)
\(\ds \leadsto \ \ \) \(\ds 3 + 996\) \(=\) \(\ds 999\)
\(\ds 15 \times 999\) \(=\) \(\ds 14 \, 985\)
\(\ds \leadsto \ \ \) \(\ds 14 + 985\) \(=\) \(\ds 999\)
\(\ds 47 \times 999\) \(=\) \(\ds 46 \, 953\)
\(\ds \leadsto \ \ \) \(\ds 46 + 953\) \(=\) \(\ds 999\)
\(\ds 57 \times 999\) \(=\) \(\ds 56 \, 943\)
\(\ds \leadsto \ \ \) \(\ds 56 + 943\) \(=\) \(\ds 999\)


Sources

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