Difference between revisions of "2021 AIME II Problems/Problem 3"
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MRENTHUSIASM (talk | contribs) (→Solution 2) |
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==Solution 2== | ==Solution 2== | ||
The expression <math>x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2</math> has cyclic symmetry. Without the loss of generality, let <math>x_1=3.</math> It follows that <math>\{x_2,x_3,x_4,x_5\}=\{1,2,4,5\}.</math> We have | The expression <math>x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2</math> has cyclic symmetry. Without the loss of generality, let <math>x_1=3.</math> It follows that <math>\{x_2,x_3,x_4,x_5\}=\{1,2,4,5\}.</math> We have | ||
− | # <math>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}</math> | + | # <math>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}.</math> |
# <math>x_2,x_3,x_4,x_5</math> are congruent to <math>1,2,1,2\pmod{3}</math> in some order. | # <math>x_2,x_3,x_4,x_5</math> are congruent to <math>1,2,1,2\pmod{3}</math> in some order. | ||
+ | |||
+ | We construct the following table for the case <math>x_1,</math> with all values in modulo <math>3:</math> | ||
+ | <cmath>\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}</cmath> | ||
<b>I am on my way. No edit please. A million thanks.</b> | <b>I am on my way. No edit please. A million thanks.</b> |
Revision as of 01:49, 23 March 2021
Problem
Find the number of permutations of numbers
such that the sum of five products
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.
We construct the following table for the case with all values in modulo
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.