Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 AIME II Problems/Problem 9"

m (→‎Solution 1)
(Undo revision 131268 by Jbala (talk))
(Tag: Undo)
 
(10 intermediate revisions by 5 users not shown)
Line 6: Line 6:
 
=== Solution 1 ===
 
=== Solution 1 ===
  
−
Ugly bashing(jk). It is actually reasonably easy to compute <math>m</math> and <math>n</math> exactly.
+
It is actually reasonably easy to compute <math>m</math> and <math>n</math> exactly.
  
 
First, note that if <math>4x+3y+2z=2009</math>, then <math>y</math> must be odd. Let <math>y=2y'-1</math>. We get <math>4x + 6y' - 3 + 2z = 2009</math>, which simplifies to <math>2x + 3y' + z = 1006</math>. For any pair of positive integers <math>(x,y')</math> such that <math>2x + 3y' < 1006</math> we have exactly one <math>z</math> such that the equality holds. Hence we need to count the pairs <math>(x,y')</math>.  
 
First, note that if <math>4x+3y+2z=2009</math>, then <math>y</math> must be odd. Let <math>y=2y'-1</math>. We get <math>4x + 6y' - 3 + 2z = 2009</math>, which simplifies to <math>2x + 3y' + z = 1006</math>. For any pair of positive integers <math>(x,y')</math> such that <math>2x + 3y' < 1006</math> we have exactly one <math>z</math> such that the equality holds. Hence we need to count the pairs <math>(x,y')</math>.  
Line 29: Line 29:
 
</cmath>
 
</cmath>
  
−
Similarly, we can compute that <math>n=82834</math>, hence <math>(m-n)\bmod 1000 = 1000\bmod 1000 = \boxed{000}</math>.
+
Similarly, we can compute that <math>n=82834</math>, hence <math>m-n = 1000 \equiv \boxed{000} \pmod{1000}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
Line 48: Line 48:
  
 
=== Solution 3 ===
 
=== Solution 3 ===
−
Perform a similar operation as in Solution 2, but only on <math>y</math>: <math>4x+3y+2z=2009</math> if and only if <math>4x+3(y-3)+2z=2000</math>. The value <math>m-n</math> is thus the additional solutions we have for this equation as opposed to the second.
+
In this solution we will perform a similar operation as in Solution 2, but only on <math>y</math>: <math>4x+3y+2z=2009</math> if and only if <math>4x+3(y-3)+2z=2000</math>. There is a one-to-one correspondence between the solutions of these two equations. Let <math>y'=y-3</math> and require <math>y'</math> to be positive as well. Then the second equation becomes <math>4x+3y'+2z=2000</math>. Notice that there are several "extra" solutions in the first equation that cannot be included in the second equation (since that would make <math>y'</math> non-positive). The value <math>m-n</math> is therefore the number of "extra" solutions.
  
−
The second equation <math>4x+3y+2z=2000</math> requires <math>y</math> to be even, so the largest solution for <math>y</math> would be <math>3y=1992</math> and the smallest <math>3y=6</math>. Now let <math>y'=y-3</math>. Note that we can make <math>3y'=3(y-3)=1992</math> without problems, and likewise with all the solutions in between; so the only additional solutions for <math>3y'</math> will be below the smallest for <math>3y</math>. Accordingly, <math>3y'</math> can equal <math>0</math> or <math>-6</math> when <math>y=3</math> or <math>y=1</math>.
+
Since <math>y'=y-3</math>, in order for <math>y'</math> to be non-positive <math>1 \leq y \leq 3</math>. However, equation (1) requires y to be odd, so we have two cases to consider: <math>y=1</math> and <math>y=3</math>. This results in the two equations <math>4x+3+2z=2009</math> and <math>4x+9+2z=2009</math>.
  
−
So there are only two equations to consider: <math>4x+0+2z=2000</math> and <math>4x-6+2z=2000</math>, or <math>2x+z=1000</math> and <math>2x+z=1003</math>.
+
<math>4x+3+2z=2009</math> simplifies to <math>2x+z=1003</math>. There is exactly one valid <math>z</math> for each <math>x</math>; <math>x</math> must be between <math>1</math> and <math>501</math> (inclusive) to obtain positive integer solutions. Therefore, there are <math>501</math> solutions in this case.  
  
−
If <math>2x+z=1000</math>, <math>x</math> must be between <math>1</math> and <math>499</math> to obtain positive integer solutions because <math>z=0</math> when <math>x=500</math>; there is exactly one valid <math>z</math> for each <math>x</math>.
+
<math>4x+9+2z=2009</math> simplifies to <math>2x+z=1000</math>. There is exactly one valid <math>z</math> for each <math>x</math>; <math>x</math> must be between <math>1</math> and <math>499</math> (inclusive) to obtain positive integer solutions. Therefore, there are <math>499</math> solutions in this case.  
  
−
If <math>2x+z=1003</math>, <math>x</math> must be between <math>1</math> and <math>501</math> to obtain positive integer solutions; there is exactly one valid <math>z</math> for each <math>x</math>.
+
Thus, <math>m-n = 501 + 499 = 1000; 1000 \equiv \boxed{000} \pmod{1000}</math>.
−
 
 
−
Therefore <math>m-n = 499 + 501 = 1000; 1000 \bmod 1000 = \boxed{000}</math>.
 
  
 
== See Also ==
 
== See Also ==

Latest revision as of 14:45, 28 December 2020

Problem

Let $m$ be the number of solutions in positive integers to the equation $4x+3y+2z=2009$, and let $n$ be the number of solutions in positive integers to the equation $4x+3y+2z=2000$. Find the remainder when $m-n$ is divided by $1000$.

Solution

Solution 1

It is actually reasonably easy to compute $m$ and $n$ exactly.

First, note that if $4x+3y+2z=2009$, then $y$ must be odd. Let $y=2y'-1$. We get $4x + 6y' - 3 + 2z = 2009$, which simplifies to $2x + 3y' + z = 1006$. For any pair of positive integers $(x,y')$ such that $2x + 3y' < 1006$ we have exactly one $z$ such that the equality holds. Hence we need to count the pairs $(x,y')$.

For a fixed $y'$, $x$ can be at most $\left\lfloor \dfrac{1005-3y'}2 \right\rfloor$. Hence the number of solutions is

\begin{align*} m & = \sum_{y'=1}^{334} \left\lfloor \dfrac{1005-3y'}2 \right\rfloor \\ & = 501 + 499 + 498 + 496 + \cdots + 6 + 4 + 3 + 1 \\ & = 1000 + 994 + \cdots + 10 + 4 \\ & = 83834 \end{align*}

Similarly, we can compute that $n=82834$, hence $m-n = 1000 \equiv \boxed{000} \pmod{1000}$.

Solution 2

We can avoid computing $m$ and $n$, instead we will compute $m-n$ directly.

Note that $4x+3y+2z=2009$ if and only if $4(x-1)+3(y-1)+2(z-1)=2000$. Hence there is an almost 1-to-1 correspondence between the positive integer solutions of the two equations. The only exceptions are the solutions of the first equation in which at least one of the variables is equal to $1$. The value $m-n$ is the number of such solutions.

If $x=1$, we get the equation $3y+2z=2005$. The variable $y$ must be odd, and it must be between $1$ and $667$, inclusive. For each such $y$ there is exactly one valid $z$. Hence in this case there are $334$ valid solutions.

If $y=1$, we get the equation $4x+2z=2006$, or equivalently $2x+z=1003$. The variable $x$ must be between $1$ and $501$, inclusive, and for each such $x$ there is exactly one valid $z$. Hence in this case there are $501$ valid solutions.

If $z=1$, we get the equation $4x+3y=2007$. The variable $y$ must be odd, thus let $y=2u-1$. We get $4x+6u=2010$, or equivalently, $2x+3u=1005$. Again, we see that $u$ must be odd, thus let $u=2v-1$. We get $2x+6v=1008$, which simplifies to $x+3v=504$. Now, we see that $v$ must be between $1$ and $167$, inclusive, and for each such $v$ we have exactly one valid $x$. Hence in this case there are $167$ valid solutions.

Finally, we must note that there are two special solutions: one with $x=y=1$, and one with $y=z=1$. We counted each of them twice, hence we have to subtract two from the total.

Therefore $m-n = 334 + 501 + 167 - 2 = 1000$, and the answer is $1000\bmod 1000 = \boxed{000}$.

Solution 3

In this solution we will perform a similar operation as in Solution 2, but only on $y$: $4x+3y+2z=2009$ if and only if $4x+3(y-3)+2z=2000$. There is a one-to-one correspondence between the solutions of these two equations. Let $y'=y-3$ and require $y'$ to be positive as well. Then the second equation becomes $4x+3y'+2z=2000$. Notice that there are several "extra" solutions in the first equation that cannot be included in the second equation (since that would make $y'$ non-positive). The value $m-n$ is therefore the number of "extra" solutions.

Since $y'=y-3$, in order for $y'$ to be non-positive $1 \leq y \leq 3$. However, equation (1) requires y to be odd, so we have two cases to consider: $y=1$ and $y=3$. This results in the two equations $4x+3+2z=2009$ and $4x+9+2z=2009$.

$4x+3+2z=2009$ simplifies to $2x+z=1003$. There is exactly one valid $z$ for each $x$; $x$ must be between $1$ and $501$ (inclusive) to obtain positive integer solutions. Therefore, there are $501$ solutions in this case.

$4x+9+2z=2009$ simplifies to $2x+z=1000$. There is exactly one valid $z$ for each $x$; $x$ must be between $1$ and $499$ (inclusive) to obtain positive integer solutions. Therefore, there are $499$ solutions in this case.

Thus, $m-n = 501 + 499 = 1000; 1000 \equiv \boxed{000} \pmod{1000}$.

See Also

2009 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_9&oldid=140845"