I need to show 11|(100a+b) if and only if 11|(a+b).
The solution:
11|(100a+b) iff 11|(100a+b-99a) so obviously 11|(a+b) and we have the solution as easily as that. However I cannot see why this is true.
2026-04-09 11:14:36.1775733276
Number theory hcf confusion
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$$\begin{align} 11 &\mid (100a+b)\\ \iff 11 &\mid (99a+a+b) \\ \iff 11 &\mid 99a+(a+b) \\ \text{and we know that }11 &\mid 99a\\ \iff11 &\mid (a+b) \\ \end{align}$$