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If $x$ is a positive integer such that $1^{x+2} + 2^{x+1} + 3^{x-1} + 4^x = 1170$, what is the value of $x$?

 Sep 14, 2020
 #1
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a=1; c=1^(a+2) + 2^(a+1) + 3^(a-1) + 4^a; if(c==1170, goto3, goto4);printc, a; a++;if(a<1000, goto1, 0)

 

x = 5

Note: Somebody will give you an algebraic solution.

 Sep 14, 2020
 #2
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$4^x < 1^{x+2}+2^{x+1}+3^{x-1}+4^x = 1170$

Now $4^6=2^{12}=4096>1170$

So $x=1,2,3,4$ or $5$.

$x=1$ doesn't work.

Take mod 3 both sides and you get

$ 1+(-1)^{x+1}+0+1 \equiv 3 (mod 3)$

So $x$ must be odd.

Therefore $x=3$ or $x=5$. Then you can calculate both cases to get $x=5$

 Sep 14, 2020
edited by Guest  Sep 14, 2020

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