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How can I prove the statement below?

$r\in \mathbb{Z}$ such that: $ 0\leq r\leq 32$

If r is the remainder of dividing the number $17*2^{635}$ by 33, then 8 divides r

Arturo Magidin
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Joaozzx2
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  • Hint: simplify your exponent with $2^5 =32=-1$ mod $33$. –  Apr 20 '20 at 21:10
  • What does "8 divides $r$" really mean when every integer is congruent with a multiple of 8 modulo 33? – Oscar Lanzi Apr 20 '20 at 21:28
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    @OscarLanzi: The question is unambiguous. The OP explicitly states that $0\le r\le 32$. – TonyK Apr 20 '20 at 21:37
  • Still awkward to me. I would have asked to find the remainder/residue explicitly instead. – Oscar Lanzi Apr 20 '20 at 22:10
  • I suspect the OP is assuming that finding the exact value of $r$ is very difficult or maybe even not possible and assumes we can only find a few properties of it. In actuality it is very easy to find that $r =16$. Exactly. – fleablood Apr 21 '20 at 05:43
  • If you do not know any number theory: $2^5=32=33-1$. $17= 33-16$. so $17*2^{635}=(33-16)(33-1)^{127}$. If we expand that out into a very long sum, every term except the very last one will be a multiple of $33$. The only term that isn't a multiple of $33$ is the last one;$(-16)\cdot(-1)= +16$. So the remainder when you divide by $33$ will be $16$. – fleablood Apr 21 '20 at 05:51

2 Answers2

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$\!\!\bmod \color{#c00}{2^{\large 5}\!+1}\!:\,\ \color{#c00}{2^{\large 5}\equiv -1}\,\Rightarrow\, 17(\color{#c00}{2^{\large 5}})^{\large 127}\!\equiv 17(\color{#c00}{-1})^{\large 127}\!\equiv -17\equiv 16\,$ via Congruence Power Rule.

Bill Dubuque
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We can calculate $r$ explicity.

Notice that $2^a\bmod 33$ is periodic with period $lcm(11-1,3-1)=10$.

Hence we need only calculate $2^5\bmod 33$ which is $-1$. So $r$ is $-17\equiv 16$, and $8$ divides $16$

Asinomás
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