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

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== Solution ==
 
== Solution ==
{{solution}}
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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>.
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Since \{ \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 \pmod(p)<math>  for all </math>k<math>. This happens when </math>p=2, 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.
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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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When </math>p=5<math> then </math>r<math> is not divisible by 5 so </math>a, b, c, d<math> are </math>1, 2, 3<math> and </math>4$, so two pairwise sums sum to 5.
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All three possible cases work so we are done.
  
 
== See Also ==
 
== See Also ==

Revision as of 13:14, 20 October 2019

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 \pmod(p)$for all$k$. This happens when$p=2, 3$or$5$.

When$ (Error compiling LaTeX. ! Missing $ inserted.)p=2$then$r$is odd, meaning$ra$,$rb$,$rc$and$rd$are all 1 mod 2, or the sum wouldn't be 2. Any pairwise sum is 2.

When$ (Error compiling LaTeX. ! Missing $ inserted.)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$ (Error compiling LaTeX. ! Missing $ inserted.)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.

See Also

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

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