Been stuck in this problem for quite a while. Apparently uses modular congruence to solve.
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Adam Hughes
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$10\equiv -1\mod 11$, so $10^n\equiv(-1)^n\mod 11$. – Nicolas FRANCOIS Nov 10 '16 at 20:22
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Write $a = \sum_{n=0}^N a_n10^n$ mod 11. Then as $10\equiv -1\mod 11$ we have
$$a\equiv\sum_{n=0}^Na_n(-1)^n\mod 11$$
which is exactly the alternating sum of its digits. If this is $0$, then by definition that's divisibility by $11$. This actually shows a stronger fact, not only is $a$ divisible by $11$ iff its alternating digital sum is, but in fact $a$ is always congruent to the alternating sum of its digits, be that sum $0$ or otherwise.
vadim123
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Adam Hughes
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Students should beware that such substitutions of an integer by a congruent integer are not generally valid. The reason they are valid above is because the expression is a composition of sums and products, and those operations enjoy this property due to the Congruence Sum and Product Rules. Generally this fails for other types of operations, e.g. powers. – Bill Dubuque Nov 10 '16 at 20:44
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Hint $ $ Radix notation has Polynomial form $\,n = d_0\! + d_1 10 + d_2 10^2\! +\cdots + d_k 10^k\! = P(10)\,$ so
${\rm mod}\ 11\!:\ \color{#c00}{10}\equiv \color{#c00}{-1}\,\Rightarrow\ n = P(\color{#c00}{10}) \equiv P(\color{#c00}{-1}) \equiv d_0 - d_1 + d_2 - \cdots + (-1)^k d_k\, $ by applying the Polynomial Congruence Rule, i.e. $\,a\equiv b\,\Rightarrow\,P(a)\equiv P(b)$.
Bill Dubuque
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