Henry Ernest Dudeney/Modern Puzzles/97 - A Common Divisor/Solution

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Modern Puzzles by Henry Ernest Dudeney: $97$

A Common Divisor
Find a common divisor for the three numbers $480 \, 608$, $508 \, 811$, and $723 \, 217$, so that the remainder shall be the same in every case.


Solution

$79$.

The remainder in each case is $51$.


Proof

Let $n$ be the divisor sought.

Let $m$ be the remainder.

We have that:

\(\ds \exists k_1 \in \Z: \, \) \(\ds 480 \, 608\) \(=\) \(\ds k_1 n + m\)
\(\ds \exists k_2 \in \Z: \, \) \(\ds 508 \, 811\) \(=\) \(\ds k_2 n + m\)
\(\ds \exists k_3 \in \Z: \, \) \(\ds 723 \, 217\) \(=\) \(\ds k_3 n + m\)
\(\ds \leadsto \ \ \) \(\ds 508 \, 811 - 480 \, 608\) \(=\) \(\ds \paren {k_2 - k_1} n\) \(\ds = 28 \, 203\)
\(\ds 723 \, 217 - 480 \, 608\) \(=\) \(\ds \paren {k_3 - k_1} n\) \(\ds = 242 \, 609\)
\(\ds 723 \, 217 - 508 \, 811\) \(=\) \(\ds \paren {k_3 - k_2} n\) \(\ds = 214 \, 406\)

Hence we see that $n$ is a common divisor for each of $28 \, 203$, $242 \, 609$ and $214 \, 406$.

So we establish the prime factors of these numbers:

\(\ds 28 \, 203\) \(=\) \(\ds 3 \times 7 \times 17 \times 79\)
\(\ds 242 \, 609\) \(=\) \(\ds 37 \times 79 \times 83\)
\(\ds 214 \, 406\) \(=\) \(\ds 2 \times 23 \times 59 \times 79\)

and it is seen by inspection that $n = 79$.

Thus we have:

\(\ds 480 \, 608\) \(=\) \(\ds 6083 \times 79 + 51\)
\(\ds 508 \, 811\) \(=\) \(\ds 6440 \times 79 + 51\)
\(\ds 723 \, 217\) \(=\) \(\ds 9154 \times 79 + 51\)

$\blacksquare$


Sources

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