Difference between revisions of "2011 IMO Problems/Problem 1"

m (Solution)
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CASE II and III:<math>k_1=</math>2, 4. Left to the reader.
 
CASE II and III:<math>k_1=</math>2, 4. Left to the reader.
  
ANSWER: <math>\{11a, a, 5a, 7a\}</math>,<math>\{a, 14a, 6a, 9a\}</math>, for any positive integer <math>a</math>.
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ANSWER: <math>\{11a, a, 5a, 7a\}</math>,<math> \{a, 11a, 19a, 29a\} </math>, for any positive integer <math>a</math>.
  
 
(Note: The above solution looks generally correct, but the actual answer should be <math>\{11a, a, 5a, 7a\}</math>,<math>\{a, 11a, 19a, 29a\}</math>. You can check that <math>\{a, 14a, 6a, 9a\}</math> doesn't actually work. -Someone who didn't write up the above solution but solved the problem in a similar way)
 
(Note: The above solution looks generally correct, but the actual answer should be <math>\{11a, a, 5a, 7a\}</math>,<math>\{a, 11a, 19a, 29a\}</math>. You can check that <math>\{a, 14a, 6a, 9a\}</math> doesn't actually work. -Someone who didn't write up the above solution but solved the problem in a similar way)

Revision as of 22:07, 19 June 2014

Problem

Given any set $A = \{a_1, a_2, a_3, a_4\}$ of four distinct positive integers, we denote the sum $a_1 +a_2 +a_3 +a_4$ by $s_A$. Let $n_A$ denote the number of pairs $(i, j)$ with $1 \leq  i < j \leq 4$ for which $a_i +a_j$ divides $s_A$. Find all sets $A$ of four distinct positive integers which achieve the largest possible value of $n_A$.

Author: Fernando Campos, Mexico

Solution

Firstly, if we order $a_1 \ge a_2 \ge a_3 \ge a_4$, we see $2(a_3 + a_4) \ge (a_1+a_2)+(a_3+a_4)  = s_A \geq 0$, so $(a_3, a_4)$ isn't a couple that satisfies the conditions of the problem. Also, $2(a_4 + a_2) = (a_4 + a_4) + (a_2 + a_2) \ge (a_4+a_3)+(a_2+a_1) = s_A \ge 0$, so again $(a_2, a_4)$ isn't a good couple. We have in total 6 couples. So $n_A \leq 6-2=4$.

We now find all sets $A$ with $n_A = 4$. If $(a,b)$ and $(c,d)$ are both good couples, and $A=\{a, b, c, d\}$, we have $a+b=c+d=s_A/2$. So WLOG $A=\{a,b,a+x,b-x\}$ with $x > 0$ and $a < b, b-x, a+x$. It's easy to see $a=a_1$ and since $(a_2, a_4),(a_3,a_4)$ are bad, all couples containing $a$ must be good. Obviously $(a,b)$ and $(a+x,b-x)$ are good ($s_A=2(a+b)$). So we have $2a+x | 2a+2b$ and $a+b-x|2a+2b \Rightarrow a+b-x|2x$.

Using the second equation, we see that if $y=a+b$, $y-x|2x \Rightarrow yk_1-xk_1=2x \Rightarrow yk_1 = x(2+k_1) \Rightarrow y=x((2+k_1)/k_1)$, for some $k_1$ a positive integer.

So now we use the first equation to get $2ak_2 + xk_2 = 2y = 2x(2+k_1)/k_1 \Rightarrow 2ak_2 = x(\frac{4+2k_1}{k_1}-k_2) \Rightarrow 2a=x(\frac{4+2k_1}{k_1k_2} - 1)$, for a natural $k_2$.

Finally, we obtain $k_1 | 4+2k-1 \Rightarrow k_1 | 4  \Rightarrow k_1=$ 1, 2 or 4. We divide in cases:

CASE I: $k_1=1$. So $y=3x$ and $2a=x((\frac{6}{k_2}) -1)$. But $a < b-x \Rightarrow 2a < y-x=2x \Rightarrow (6/k_2) - 1 < 2 \Rightarrow k_2 > 2 \Rightarrow k_2 =$3, 4,5 or 6. $k_2=6$ implies $a=0$, impossible. $a=x$ when $k_2=3$. We easily see $b=3x=3x$ and $A=\{x, 3x, 3x-x, 2x\}$, impossible since $3x-x=2x$. When $k_2=4$, $a=x/4=y/12$, and we get $\{11a, a, 5a, 7a\}$.Uf $k_2=5$, $a=x/5=y/15$ and we get $\{a, 14a, 6a, 9a\}$.

CASE II and III:$k_1=$2, 4. Left to the reader.

ANSWER: $\{11a, a, 5a, 7a\}$,$\{a, 11a, 19a, 29a\}$, for any positive integer $a$.

(Note: The above solution looks generally correct, but the actual answer should be $\{11a, a, 5a, 7a\}$,$\{a, 11a, 19a, 29a\}$. You can check that $\{a, 14a, 6a, 9a\}$ doesn't actually work. -Someone who didn't write up the above solution but solved the problem in a similar way)

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
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