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>. | ||
− | Since \{ \frac{ra}p \}+\{ \frac{rb}p \}+\{ \frac{rc}p \}+\{ \frac{rd}p \} =2<math>, then we see that they have to represent mods < | + | 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 \pmod(p)</math> for all <math>k</math>. This happens when <math>p=2, 3</math> or <math>5</math>. |
− | When < | + | 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 < | + | 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 < | + | 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</math>, so two pairwise sums sum to 5. |
All three possible cases work so we are done. | All three possible cases work so we are done. |
Revision as of 14:15, 20 October 2019
Problem
Let be a prime and let be integers not divisible by , such that for any integer not divisible by . Prove that at least two of the numbers , , , , , are divisible by . (Note: denotes the fractional part of .)
Solution
We see that means that . Now, since does nto divide and is prime, their GCD is 1 so .
Since , then we see that they have to represent mods , and thus, our possible values of are all such that for all . This happens when or .
When then is odd, meaning , , and are all 1 mod 2, or the sum wouldn't be 2. Any pairwise sum is 2.
When then is not divisible by 3, thus two are , and the other two are . Thus, four pairwise sums sum to 3.
When then is not divisible by 5 so are and , so two pairwise sums sum to 5.
All three possible cases work so we are done.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.