Difference between revisions of "2011 IMO Problems/Problem 1"
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==Problem== | ==Problem== | ||
− | Given any set <math>A = \{a_1, a_2, a_3, a_4\}</math> of four distinct positive integers, we denote the sum <math>a_1+a_2+a_3+a_4</math> by <math>s_A</math>. Let <math>n_A</math> denote the number of pairs <math>(i,j)</math> with <math>1 \leq i < j \leq 4</math> for which <math>a_i+a_j</math> divides <math>s_A</math>. Find all sets <math>A</math> of four distinct positive integers which achieve the largest possible value of <math>n_A</math>. | + | Given any set <math>A = \{a_1, a_2, a_3, a_4\}</math> of four distinct positive integers, we denote the sum <math>a_1 +a_2 +a_3 +a_4</math> by <math>s_A</math>. Let <math>n_A</math> denote the number of pairs <math>(i, j)</math> with <math>1 \leq i < j \leq 4</math> for which <math>a_i +a_j</math> divides <math>s_A</math>. Find all sets <math>A</math> of four distinct positive integers which achieve the largest possible value of <math>n_A</math>. |
+ | |||
+ | ''Author: Fernando Campos, Mexico'' | ||
==Solution== | ==Solution== |
Revision as of 14:02, 21 June 2012
Problem
Given any set of four distinct positive integers, we denote the sum
by
. Let
denote the number of pairs
with
for which
divides
. Find all sets
of four distinct positive integers which achieve the largest possible value of
.
Author: Fernando Campos, Mexico
Solution
Firstly, if we order , we see
, so
isn't a couple that satisfies the conditions of the problem. Also,
, so again
isn't a good couple. We have in total 6 couples. So
.
We now find all sets with
. If
and
are both good couples, and
, we have
.
So WLOG
with
and
. It's easy to see
and since
are bad, all couples containing
must be good. Obviously
and
are good (
). So we have
and
.
Using the second equation, we see that if ,
, for some
a positive integer.
So now we use the first equation to get , for a natural
.
Finally, we obtain 1, 2 or 4. We divide in cases:
CASE I: .
So
and
. But
3, 4,5 or 6.
implies
, impossible.
when
. We easily see
and
, impossible since
. When
,
, and we get
.Uf
,
and we get
.
CASE II and III:2, 4. Left to the reader.
ANSWER: ,
, for any positive integer
.
See Also
2011 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |