2021 AIME II Problems/Problem 3

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 $x_2,x_3,x_4,x_5$ are $1,2,4,5$ in some order. In modulo $3,$ they are $1,2,1,2$ in some order.

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

~MRENTHUSIASM

2021 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2021 AIME II Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See also