Find smallest number m such that $9^{32} + 19^{433} + m$ is divisible by $4$

Question

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.

Answer 1

$9\equiv 1\pmod4$, 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 $4$.

Answer 2

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.)

Answer 3

Hint $\rm\,\ mod\ n\!:\ (an\!+\!1)^j + (bn\!-\!1)^{2k+1}\equiv 1^j + (-1)^{2k+1}\equiv 1-1\equiv 0$

Answer 4

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.

Answer 5

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!