2021 AIME II Problems/Problem 3
Problem
Find the number of permutations 
of numbers 
such that the sum of five products ![\[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\]](http://latex.artofproblemsolving.com/5/f/8/5f85b1e5c0060aca6f470f54f8117ac7500be017.png)
Solution 1
Since 
is one of the numbers, a product with a in it is automatically divisible by , so WLOG 
, we will multiply by 
afterward since any of 
would be , after some cancelation we see that now all we need to find is the number of ways that 
is divisible by , since 
is never divisible by , now we just need to find the number of ways 
is divisible by , after some calculation you will see that there are 
ways to choose 
and 
in this way. So the desired answer is 
.
~ math31415926535
Solution 2
The expression 
has cyclic symmetry. Without the loss of generality, let 
It follows that 
We have
are congruent to 
in some order.
are congruent to 
in some order.We construct the following table for the case 
with all values in modulo 
![\[\begin{array}{c|c|c|c|c|c} \boldsymbol{x_2} & \boldsymbol{x_3} & \boldsymbol{x_4} & \boldsymbol{x_5} & \boldsymbol{x_2x_3x_4 + x_3x_4x_5} & \textbf{Valid?} \\ \hline & & & & & \\ [-2ex] 1 & 1 & 2 & 2 & 0 & \checkmark \\ 1 & 2 & 1 & 2 & 0 & \checkmark \\ 1 & 2 & 2 & 1 & 2 & \\ 2 & 1 & 1 & 2 & 1 & \\ 2 & 1 & 2 & 1 & 0 & \checkmark \\ 2 & 2 & 1 & 1 & 0 & \checkmark \end{array}\]](http://latex.artofproblemsolving.com/1/b/1/1b1b39fad7e5e6f54f1a1a4054380e785249ff28.png)
I am on my way. No edit please. A million thanks.
~MRENTHUSIASM
Solution 3
See also
| 2021 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
