Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999
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
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $999$
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $142,857$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $999$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $142,857$