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Difference between revisions of "1999 USAMO Problems/Problem 3"

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We see that <math>\{\frac{ra+rb+rc+rd}{p}\}=0</math> means that <math>p|r(a+b+c+d)</math>. Now, since <math>p</math> does nto divide <math>r</math> and <math>p</math> is prime, their GCD is 1 so <math>p|a+b+c+d</math>.
 
We see that <math>\{\frac{ra+rb+rc+rd}{p}\}=0</math> means that <math>p|r(a+b+c+d)</math>. Now, since <math>p</math> does nto divide <math>r</math> and <math>p</math> is prime, their GCD is 1 so <math>p|a+b+c+d</math>.
  
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Since <math>\{ \frac{ra}{p} \}+\{ \frac{rb}{p} \}+\{ \frac{rc}{p} \}+\{ \frac{rd}{p} \} =2</math>, then we see that they have to represent mods <math>\mod p</math>, and thus, our possible values of <math>p</math> are all such that <math>k^4 \equiv 1 \mod(p)</math>  for all <math>k</math>. This happens when <math>p=2, 3</math> or <math>5</math>.
+
Since <math>\{ \frac{ra}{p} \}+\{ \frac{rb}{p} \}+\{ \frac{rc}{p} \}+\{ \frac{rd}{p} \} =2</math>, then we see that they have to represent mods <math>\mod p</math>, and thus, our possible values of <math>p</math> are all such that <math>k^4 \equiv 1 \mod(p)</math>  for all <math>k</math>. This happens when <math>p=3</math> or <math>5</math>.  
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When <math>p=2</math> then <math>r</math> is odd, meaning <math>ra</math>, <math>rb</math>, <math>rc</math> and <math>rd</math> are all 1 mod 2, or the sum wouldn't be 2. Any pairwise sum is 2.
 
  
 
When <math>p=3</math> then <math>r</math> is not divisible by 3, thus two are <math>1</math>, and the other two are <math>2</math>. Thus, four pairwise sums sum to 3.
 
When <math>p=3</math> then <math>r</math> is not divisible by 3, thus two are <math>1</math>, and the other two are <math>2</math>. Thus, four pairwise sums sum to 3.
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All three possible cases work so we are done.
 
All three possible cases work so we are done.
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 +
(This solution makes absolutely no sense. Why does <math>k^4\equiv 1</math>??)
  
 
== See Also ==
 
== See Also ==

Revision as of 10:50, 10 July 2020

Problem

Let $p > 2$ be a prime and let $a,b,c,d$ be integers not divisible by $p$, such that \[\left\{ \dfrac{ra}{p} \right\} + \left\{ \dfrac{rb}{p} \right\} + \left\{ \dfrac{rc}{p} \right\} + \left\{ \dfrac{rd}{p} \right\} = 2\] for any integer $r$ not divisible by $p$. Prove that at least two of the numbers $a+b$, $a+c$, $a+d$, $b+c$, $b+d$, $c+d$ are divisible by $p$. (Note: $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x$.)

Solution

We see that $\{\frac{ra+rb+rc+rd}{p}\}=0$ means that $p|r(a+b+c+d)$. Now, since $p$ does nto divide $r$ and $p$ is prime, their GCD is 1 so $p|a+b+c+d$.

Since $\{ \frac{ra}{p} \}+\{ \frac{rb}{p} \}+\{ \frac{rc}{p} \}+\{ \frac{rd}{p} \} =2$, then we see that they have to represent mods $\mod p$, and thus, our possible values of $p$ are all such that $k^4 \equiv 1 \mod(p)$ for all $k$. This happens when $p=3$ or $5$.

When $p=3$ then $r$ is not divisible by 3, thus two are $1$, and the other two are $2$. Thus, four pairwise sums sum to 3.

When $p=5$ then $r$ is not divisible by 5 so $a, b, c, d$ are $1, 2, 3$ and $4$, so two pairwise sums sum to 5.

All three possible cases work so we are done.

(This solution makes absolutely no sense. Why does $k^4\equiv 1$??)

See Also

1999 USAMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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