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

(→‎Solution 2)
(→‎Solution 3)
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==Solution 3==
 
==Solution 3==
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WLOG, let <math>x_{3} = 3</math>
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So, the terms <math>x_{1}x_{2}x_{3}, x_{2}x_{3}x_{4},x_{3}x_{4}x_{5}</math> are divisible by <math>3</math>.
 +
 +
We are left with <math>x_{4}x_{5}x_{1}</math> and <math>x_{5}x_{1}x_{2}</math>.
 +
We need <math>x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} \equiv 0 (mod 3)</math>
 +
The only way is when They are <math>(+1,-1)</math> or <math>(-1, +1)</math> (mod 3)
 +
 +
The numbers left with us are <math>1,2,4,5</math> which are <math>+1,-1,+1,-1</math> (mod <math>3</math>) respectively.
 +
 +
<math>+1 (of x_{4}x_{5}x_{1} or x_{5}x_{1}x_{2}) = +1 \cdot +1 \cdot +1 or +1 (of x_{4}x_{5}x_{1} or x_{5}x_{1}x_{2}) = -1 \cdot -1 \cdot +1</math>.
 +
 +
<math>-1 (of x_{4}x_{5}x_{1} or x_{5}x_{1}x_{2}) = -1 \cdot -1 \cdot -1 or -1 (of x_{4}x_{5}x_{1} or x_{5}x_{1}x_{2}) = -1 \cdot +1 \cdot +1</math>
 +
 +
But, as we have just two <math>+1's</math> and two <math>-1's</math>,
 +
Hence, We will have to take <math>+1 = +1 \cdot -1 \cdot -1</math> and <math>-1 = -1 \cdot +1 \cdot +1</math>
 +
Among these two, we have a <math>+1</math> and <math>-1</math> in common, i.e. <math>(x_{5}, x_{1}) = (+1, -1) or (-1, +1)</math> (because <math>x_{1}</math> and <math>x_{5}</math>
 +
are common in <math>x_{4}x_{5}x_{1}</math> and <math>x_{5}x_{1}x_{2}</math> )
 +
 +
So, <math>(x_{5}, x_{1}) \in {(1,2), (1,5), (4,2), (4,5), (2,1), (5,1), (2,4), (5,4)}</math> i.e. <math>8</math> values.
 +
 +
For each value of <math>(x_{5}, x_{1})</math> we get <math>2</math> values for <math>(x_{2}, x_{4})</math>
 +
Hence, in total, we have <math>8 \times 2 = \boxed{16}</math> ways.
 +
 +
-Arnav Nigam
  
 
==See also==
 
==See also==
 
{{AIME box|year=2021|n=II|num-b=2|num-a=4}}
 
{{AIME box|year=2021|n=II|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:59, 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

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

We construct the following table for the case $x_1,$ with all values in modulo $3:$ \[\begin{array}{c|c|c|c|c|c} \boldsymbol{x_2} & \boldsymbol{x_3} & \boldsymbol{x_4} & \boldsymbol{x_5} & \boldsymbol{x_2x_3x_4 + x_3x_4x_5} & \textbf{Valid?} \\ \hline & & & & & \\ [-2ex] 1 & 1 & 2 & 2 & 0 & \checkmark \\ 1 & 2 & 1 & 2 & 0 & \checkmark \\ 1 & 2 & 2 & 1 & 2 & \\ 2 & 1 & 1 & 2 & 1 & \\ 2 & 1 & 2 & 1 & 0 & \checkmark \\ 2 & 2 & 1 & 1 & 0 & \checkmark \end{array}\]

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

~MRENTHUSIASM

Solution 3

WLOG, let $x_{3} = 3$ So, the terms $x_{1}x_{2}x_{3}, x_{2}x_{3}x_{4},x_{3}x_{4}x_{5}$ are divisible by $3$.

We are left with $x_{4}x_{5}x_{1}$ and $x_{5}x_{1}x_{2}$. We need $x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} \equiv 0 (mod 3)$ The only way is when They are $(+1,-1)$ or $(-1, +1)$ (mod 3)

The numbers left with us are $1,2,4,5$ which are $+1,-1,+1,-1$ (mod $3$) respectively.

$+1 (of x_{4}x_{5}x_{1} or x_{5}x_{1}x_{2}) = +1 \cdot +1 \cdot +1 or +1 (of x_{4}x_{5}x_{1} or x_{5}x_{1}x_{2}) = -1 \cdot -1 \cdot +1$.

$-1 (of x_{4}x_{5}x_{1} or x_{5}x_{1}x_{2}) = -1 \cdot -1 \cdot -1 or -1 (of x_{4}x_{5}x_{1} or x_{5}x_{1}x_{2}) = -1 \cdot +1 \cdot +1$

But, as we have just two $+1's$ and two $-1's$, Hence, We will have to take $+1 = +1 \cdot -1 \cdot -1$ and $-1 = -1 \cdot +1 \cdot +1$ Among these two, we have a $+1$ and $-1$ in common, i.e. $(x_{5}, x_{1}) = (+1, -1) or (-1, +1)$ (because $x_{1}$ and $x_{5}$ are common in $x_{4}x_{5}x_{1}$ and $x_{5}x_{1}x_{2}$ )

So, $(x_{5}, x_{1}) \in {(1,2), (1,5), (4,2), (4,5), (2,1), (5,1), (2,4), (5,4)}$ i.e. $8$ values.

For each value of $(x_{5}, x_{1})$ we get $2$ values for $(x_{2}, x_{4})$ Hence, in total, we have $8 \times 2 = \boxed{16}$ ways.

-Arnav Nigam

See also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AIME Problems and Solutions

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