I wanted to post this as a problem but decided to put it as a note as I put an extended version of this.

Two numbers, 823519 and 274658, such that if the first number is divided by \(x\), the remainder would be three times the remainder obtained by dividing the second number by \(x\). Find and prove that \(x\) has only one value.

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## Comments

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TopNewest\[274658\equiv r\pmod x\stackrel{\times 3}\implies 823974\equiv 3r\pmod x\]

\[\implies 823974-823519\equiv3r-3r\equiv 455\equiv 0\pmod x\implies x\mid 5\cdot 7\cdot 13\]

There's now a limited amount of \(x\)'s that could work - only the factors of \(5\cdot 7\cdot 13\), which after bashing (all the \(7\) possibilities) gives \(\boxed{91}\) as the only solution.

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