Difference between revisions of "1999 USAMO Problems/Problem 3"
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== Solution == | == Solution == | ||
− | 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 | + | We see that <math>\biggl\{\frac{ra+rb+rc+rd}{p}\biggr\}=0</math> means that <math>p|r(a+b+c+d)</math>. Now, since <math>p</math> does not divide <math>r</math> and <math>p</math> is prime, their GCD is 1 so <math>p\mathrel{|}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>\biggl\{\frac{ra}{p}\biggr\}+\biggl\{\frac{rb}{p}\biggr\}+\biggl\{\frac{rc}{p}\biggr\}+\biggl\{\frac{rd}{p}\biggr\}=2</math>, then we see that they have to represent mods <math>\bmod\medspace 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> that are relatively prime to <math>p</math>. This happens when <math>p=3</math> or <math>5</math>. |
− | 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. | |
− | + | (This solution makes absolutely no sense. Why is <math>k^4\equiv 1</math>? And how do we know that only <math>3</math> and <math>5</math> work!?) | |
== See Also == | == See Also == |
Latest revision as of 11:17, 18 March 2023
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 not 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
that are relatively prime to
. This happens when
or
.
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.
(This solution makes absolutely no sense. Why is ? And how do we know that only
and
work!?)
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.