We went over this in class awhile ago, but I can't seem to figure out how to solve it. Obviously you can do it exhaustively with a supercomputer, but that doesn't seem practical when I know there's a simplistic way to solve it.
Find smallest number m such that $9^{32} + 19^{433} + m$ is divisible by $4$
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0It’s a good idea to make your question self-contained, rather than having essential parts only in the title. – 2012-11-28
5 Answers
$9\equiv 1\pmod4$, so every power of $9$ is also congruent to $1\bmod 4$; in particular, $9^{32}\equiv 1\pmod 4$. $19\equiv -1\pmod 4$, so $19^{433}\equiv(-1)^{433}\equiv -1\pmod 4$, since $(-1)^{433}=-1$. Thus, $9^{32}+19^{433}\equiv 1+(-1)\equiv 0\pmod 4\;,$ meaning that $9^{32}+19^{433}$ is divisible by $4$. The smallest non-negative $m$ such that $9^{32}+19^{433}+m$ is divisible by $4$ is therefore $0$, and the smallest positive $m$ is $4$.
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0@DougSmith $9 \neq 1 ~ \mathrm{mod} ~ 4$, however $9 \equiv 1 ~ \mathrm{mod} ~ 4$ (congruence) - it's important to distinguish them. – 2012-11-28
Since you've gotten the mathematical answer already, I'd like to point out that it's not a particularly hard problem computationally (and certainly doesn't require a supercomputer). It can be computed in GAP as follows:
gap> 9^32+19^433; 501544200119392562625247277746089908092336003464306403005859888420683628454522\ 294504197711028243108230887275382154581016491417414644491962851915431414455652\ 288045095893781599660030753306640524923972275437754069448537906581323068453513\ 432352483965776742110824103821944465300751947471969610305958014477196847328310\ 211743932024655151477385264075258820700008154477693517186089623494722993687602\ 585620203537058139081722096639662034404319726334084194425480243598759870559077\ 000920199390664370931533991434513332012987998517097840035144875728681773054413\ 22534740 gap> last mod 4; 0
Since every 4-th integer is divisible by 4, and $9^{32}+19^{433}$ is divisible by 4, we can deduce that $9^{32}+19^{433}+m$ is divisible by $4$ if and only if $m$ is divisible by $4$.
This was computed on my home computer, and took less than a microsecond to compute (it took me significantly longer to type 9^32+19^433;
). You could similarly compute it using a zillion other computer algebra systems, or even Wolfram|Alpha.
(Note: while I'm not advocating solely relying on computers to solve mathematical problems, I think it's useful to know what the computer will be able to do.)
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0@Thomas: thanks for pointing this out. Just in case, GAP could do it with `PowerModInt`, e.g. `PowerModInt(19,433,4);` – 2013-09-03
Hint $\rm\,\ mod\ n\!:\ (an\!+\!1)^j + (bn\!-\!1)^{2k+1}\equiv 1^j + (-1)^{2k+1}\equiv 1-1\equiv 0$
I presume “the smallest number m” here is intended in the sense of “the smallest non-negative integer m”. Assuming this, I’d suggest:
first, show that for any number n, “what’s the smallest number m such that n + m is divisible by 4” is equivalent to a slightly different question about n, in a more standard form.
secondly, use techniques about powers (which you’ve hopefully seen) to answer that question.
It seems you can do this simply by modding out: $ 9^{32}\equiv 81^{16}\equiv 1^{16}\equiv 1 \mod 4 $ $ 19^{433}\equiv (-1)^{433}\equiv -1 \mod 4 $
Adding these, we see $9^{32}+19^{433}$ is divisible by four, so $m=0$ is valid. I assume you mean the smallest absolute value, as you can just make negative multiples of four. Hope that's what you were looking for!
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0Oops. I have no clue how how a got 20. Luckily, it works, and I'll fix that now. @ Doug, $19\equiv 3\equiv -1$ since modular arithmetic can easily be applied in negative numbers, and it is clear $19-5\cdot 4=-1$. – 2012-11-28