Difference between revisions of "2021 AIME II Problems/Problem 3"

Revision as of 00:41, 23 March 2021

Problem

Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2$

Solution 1

Since $3$ is one of the numbers, a product with a $3$ in it is automatically divisible by $3$ , so WLOG $x_3=3$ , we will multiply by $5$ afterward since any of $x_1, x_2, ..., x_5$ would be $3$ , after some cancelation we see that now all we need to find is the number of ways that $x_5x_1(x_4+x_2)$ is divisible by $3$ , since $x_5x_1$ is never divisible by $3$ , now we just need to find the number of ways $x_4+x_2$ is divisible by $3$ , after some calculation you will see that there are $16$ ways to choose $x_1, x_2, x_4,$ and $x_5$ in this way. So the desired answer is $16 \times 5=\boxed{080}$ .

~ math31415926535

Solution 2

The expression $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2$ has cyclic symmetry. Without the loss of generality, let $x_1=3.$ It follows that $\{x_2,x_3,x_4,x_5\}=\{1,2,4,5\}.$ We have

$x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\equiv x_2x_3x_4 + x_3x_4x_5\pmod{3}$
$x_2,x_3,x_4,x_5$ are congruent to $1,2,1,2\pmod{3}$ in some order.

I am on my way. No edit please. A million thanks.

~MRENTHUSIASM

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 ~MRENTHUSIASM
+==Solution 3==
 ==See also==
 {{AIME box|year=2021|n=II|num-b=2|num-a=4}}
 {{MAA Notice}}

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Difference between revisions of "2021 AIME II Problems/Problem 3"

Revision as of 00:41, 23 March 2021

Contents

Problem

Solution 1

Solution 2

Solution 3

See also