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An elaboration on this is that when factoring smallish numbers of the form \(n = 10k+1\) you will often need to test for divisibility by 7 and then by 13 in succession. But \( 7\cdot 13 = 91\) so instead of considering \( n\), consider \[𝑛−91=10k−90=10\cdot (k-9)\] and do both at once.

For example, instead of dividing \(5031\) by \(7\) and then by \(13\), consider \(503-9 = 494\) instead. This is evidently not divisible by \( 7\), but for \(13 \) you can see from \( 494 + 26 = 520 \) that it is a multiple of \(13\), and therefore so too was \(5031\).