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I have come across this question, and need a bit of help understanding it.

For integers $a$ and $b$, with $a \geq 0$ and $0 \leq b \leq 99$, the numbers $M$ and $N$ are such that
$$M = 100a + b$$ $$N = a - 9b$$ (i) By considering the number $M+2N$, show that $17 | M$ if and only if $17 | N$.
(ii) Demonstrate step-by-step how an algorithm based on the result of part (i) can be used to show that $2 058 376 813 901$ is a multiple of 1 7.

For the first one, it's easy to see that $M+2N = 102a-17b =17(6a-b)$, though I'm not sure how to continue. I feel as if I am missing something completely.

For the second question, I was told to do as follows: $$2 058 3768139\underline{01} \rightarrow 20583768139 - (9\times 1) = 20583768130$$ $$205837681\underline{30} \rightarrow 205837681 - (9 \times 30) = 205837411$$ $$2058374\underline{11} \rightarrow 2058374 - (9 \times 11) = 2058275 $$ $$20582\underline{75} \rightarrow 20582 - (9 \times 75) = 19907$$ $$199\underline{07} \rightarrow 199 - (9 \times 7) = 136$$ $$136 = 17 \times 8$$

However, I do not understand this solution, most likely due to the fact I lack the understanding for part (i).

Bill Dubuque
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lulu
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  • Special case (radix $=\color{#0a0}{100})$ of the general test in this answer in the linked dupe, using $\bmod 17!:,\ \color{#0a0}{100\equiv -2}\Rightarrow 1/\color{#0a0}{100}\equiv 18/(\color{#0a0}{-2})\equiv \color{#c00}{-9},,$ so $,17\mid \color{#0a0}{100}a+b\iff 17\mid a\color{#c00}{-9}b\ \ $ – Bill Dubuque Jul 14 '21 at 21:55

1 Answers1

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You're on the right track for (1). We have

$$M+2N=17(6a-b)$$

Suppose $17 \mid N$. Therefore $N=17k$ for some $k \in \mathbb{Z}$. Substitution gives us

$$M+2(17k)=17(6a-b) \implies M = 17(6a-b-2k) \implies 17 \mid M$$

Now suppose $17\mid M$. Therefore $M=17q$ for some $q \in \mathbb{Z}$. Substitution gives us

$$17q+2N=17(6a-b) \implies 2N = 17(6a-b-q)$$

$17$ is prime, so according to Euclid's lemma, if $17 \mid 2N$ then at least one of $17 \mid 2$ or $17 \mid N$ is true. Since $17 \nmid 2$, we have $17 \mid N$. $\blacksquare$

For (2), we can see that

$$2058376813901=100(20583768139)+1$$

so let $M=2058376813901$. We've shown that $a$ must be $20583768139$ and $b$ must be $9(1)=9$ so we have $N=20583768139-9=20583768130$. Then you follow the solution you were provided to prove that $17 \mid N$ and use the logic from (1) and you're done.

Kman3
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