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Difference between revisions of "2012 USAJMO Problems/Problem 5"

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First we'll show that <math>S \geq 502</math>, then we'll find an example <math>(a, b)</math> that have <math>f(a, b)=502</math>.
 
First we'll show that <math>S \geq 502</math>, then we'll find an example <math>(a, b)</math> that have <math>f(a, b)=502</math>.
  
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Let <math>x_k</math> be the remainder when <math>ak</math> is divided by 2012, and let <math>y_k</math> be defined similarly for <math>bk</math>. First, we know that, if <math>x_k > y_k</math> and <math>bk \nequiv 0 \pmod 2012</math>, then <math>x_{2012-k} \equiv a(2012-k) \equiv 2012-ak \equiv 2012-x_k \pmod 2012</math> and <math>y_{2012-k} \equiv 2012-y_k \pmod 2012</math>. This implies that, since <math>2012 - x_k \neq 0</math> and <math>2012 -y_k \neq 0</math>, <math>x_{2012-k} < y_{2012-k}</math>. Similarly, if <math>x_k < y_k</math> and <math>ak \nequiv 0 \pmod 2012</math> then <math>x_{2012-k} > y_{2012-k}</math>, establishing a one-to-one correspondence between the number of <math>k</math> such that <math>x_k < y_k</math>. Thus, if <math>n</math> is the number of <math>k</math> such that <math>x_k \neq y_k</math> and <math>y_k \neq 0 \neq x_k </math>, then <math>S \geq \frac{1}{2}n</math>. Now I'll show that <math>n \geq 1004</math>.
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Let <math>x_k</math> be the remainder when <math>ak</math> is divided by 2012, and let <math>y_k</math> be defined similarly for <math>bk</math>. First, we know that, if <math>x_k > y_k >0</math>, then <math>x_{2012-k} \equiv a(2012-k) \equiv 2012-ak \equiv 2012-x_k \pmod 2012</math> and <math>y_{2012-k} \equiv 2012-y_k \pmod 2012</math>. This implies that, since <math>2012 - x_k \neq 0</math> and <math>2012 -y_k \neq 0</math>, <math>x_{2012-k} < y_{2012-k}</math>. Similarly, if <math>0< x_k < y_k</math> then <math>x_{2012-k} > y_{2012-k}</math>, establishing a one-to-one correspondence between the number of <math>k</math> such that <math>x_k < y_k</math>. Thus, if <math>n</math> is the number of <math>k</math> such that <math>x_k \neq y_k</math> and <math>y_k \neq 0 \neq x_k </math>, then <math>S \geq \frac{1}{2}n</math>. Now I'll show that <math>n \geq 1004</math>.
  
 
If <math>gcd(k, 2012)=1</math>, then I'll show you that <math>x_k \neq y_k</math>. This is actually pretty clear; assume that's not true and set up a congruence relation:
 
If <math>gcd(k, 2012)=1</math>, then I'll show you that <math>x_k \neq y_k</math>. This is actually pretty clear; assume that's not true and set up a congruence relation:

Revision as of 17:19, 12 June 2019

Problem

For distinct positive integers $a$, $b < 2012$, define $f(a,b)$ to be the number of integers $k$ with $1 \le k < 2012$ such that the remainder when $ak$ divided by 2012 is greater than that of $bk$ divided by 2012. Let $S$ be the minimum value of $f(a,b)$, where $a$ and $b$ range over all pairs of distinct positive integers less than 2012. Determine $S$.

Solutions

Solution 1

First we'll show that $S \geq 502$, then we'll find an example $(a, b)$ that have $f(a, b)=502$.

Let $x_k$ be the remainder when $ak$ is divided by 2012, and let $y_k$ be defined similarly for $bk$. First, we know that, if $x_k > y_k >0$, then $x_{2012-k} \equiv a(2012-k) \equiv 2012-ak \equiv 2012-x_k \pmod 2012$ and $y_{2012-k} \equiv 2012-y_k \pmod 2012$. This implies that, since $2012 - x_k \neq 0$ and $2012 -y_k \neq 0$, $x_{2012-k} < y_{2012-k}$. Similarly, if $0< x_k < y_k$ then $x_{2012-k} > y_{2012-k}$, establishing a one-to-one correspondence between the number of $k$ such that $x_k < y_k$. Thus, if $n$ is the number of $k$ such that $x_k \neq y_k$ and $y_k \neq 0 \neq x_k$, then $S \geq \frac{1}{2}n$. Now I'll show that $n \geq 1004$.

If $gcd(k, 2012)=1$, then I'll show you that $x_k \neq y_k$. This is actually pretty clear; assume that's not true and set up a congruence relation: \[ak \equiv bk \pmod 2012\] Since $k$ is relatively prime to 2012, it is invertible mod 2012, so we must have $a \equiv b \pmod 2012$. Since 0 < a, b <2012$, this means$a=b$, which the problem doesn't allow, thus contradiction, and$x_k \neq y_k$. Additionally, if$gcd(k, 2012)=1$, then$x_k \neq 0 \neq y_k$, then based on what we know about$n$from the previous paragraph,$n$is at least as large as the number of k relatively prime to 2012. Thus,$n \geq \phi(2012) = \phi(503*4) = 1004$. Thus,$S \geq 502$.

To show 502 works, consider$ (Error compiling LaTeX. Unknown error_msg)(a, b)=(1006, 2)$. For all even$k$we have$x_k=0$, so it doesn't count towards$f(1006, 2)$. Additionally, if$k = 503, 503*3$then$x_k = y_k = 1006$, so the only number that count towards$f(1006, 2)$are the odd numbers not divisible by 503. There are 1004 such numbers. However, for all such odd k not divisible by 503 (so numbers relatively prime to 2012), we have$x_k \neq 0 \neq y_k$and$2012-k$is also relatively prime to 2012. Since under those conditions exactly one of$x_k > y_k$and$x_{2012-k} > y_{2012-k}$is true, we have at most 1/2 of the 1004 possible k actually count to$f(1006, 2)$, so$\frac{1004}{2} = 502 \geq f(1006, 2) \geq S \geq 502$, so$S=502$. === Solution 2 ===

Let$ (Error compiling LaTeX. Unknown error_msg)ak \equiv r_{a} \pmod{2012}$and$bk \equiv r_{b} \pmod{2012}$. Notice that this means$a(2012 - k) \equiv 2012 - r_{a} \pmod{2012}$and$b(2012 - k) \equiv 2012 - r_{b} \pmod{2012}$. Thus, for every value of$k$where$r_{a} > r_{b}$, there is a value of$k$where$r_{b} > r_{a}$. Therefore, we merely have to calculate$\frac{1}{2}$times the number of values of$k$for which$r_{a} \neq r_{b}$and$r_{a} \neq 0$.


However, the answer is NOT$ (Error compiling LaTeX. Unknown error_msg)\frac{1}{2}(2012) = 1006$! This is because we must count the cases where the value of$k$makes$r_{a} = r_{b}$or where$r_{a} = 0$.


So, let's start counting.


If$ (Error compiling LaTeX. Unknown error_msg)k$is even, we have either$a \equiv 0 \pmod{1006}$or$a - b \equiv 0 \pmod{1006}$. So,$a = 1006$or$a = b + 1006$. We have$1005$even values of$k$(which is all the possible even values of$k$, since the two above requirements don't put any bounds on$k$at all).


If$ (Error compiling LaTeX. Unknown error_msg)k$is odd, if$k = 503$or$k = 503 \cdot 3$, then$a \equiv 0 \pmod{4}$or$a \equiv b \pmod{4}$. Otherwise,$ak \equiv 0 \pmod{2012}$or$ak \equiv bk \pmod{2012}$, which is impossible to satisfy, given the domain$a, b < 2012$. So, we have$2$values of$k$.


In total, we have$ (Error compiling LaTeX. Unknown error_msg)2 + 1005 = 1007$values of$k$which makes$r_{a} = r_{b}$or$r_{a} = 0$, so there are$2011 - 1007 = 1004$values of$k$for which$r_{a} \neq r_{b}$and$r_{a} \neq 0$. Thus, by our reasoning above, our solution is$\frac{1}{2} \cdot 1004 = \boxed{502}$.


Solution by$ (Error compiling LaTeX. Unknown error_msg)\textbf{\underline{Invoker}}$=== Solution 3 === The key insight in this problem is noticing that when$ak$is higher than$bk$,$a(2012-k)$is lower than$ b(2012-k)$, except at$2 \pmod{4}$residues*. Also, they must be equal many times.$2012=2^2*503$. We should have multiples of$503$. After trying all three pairs and getting$503$as our answer, we win. But look at the$2\pmod{4}$idea. What if we just took$2$and plugged it in with$1006$? We get$502$. 

--[[User:Va2010|Va2010]] 11:12, 28 April 2012 (EDT)va2010

=== Solution 4 === Say that the problem is a race track with$ (Error compiling LaTeX. Unknown error_msg)2012$spots. To intersect the most, we should get next to each other a lot so the negation is high. As$2012=2^2*503$, we intersect at a lot of multiples of$503$.

See also

2012 USAJMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

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