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

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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.
 
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.
  
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 will be below the smallest. Accordingly, <math>y'</math> can equal <math>0</math> or <math>-6</math> when <math>y=3</math> or <math>y=1</math>.
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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>.
  
 
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>.
 
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>.

Revision as of 12:44, 31 March 2013

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

Brute force. 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 $(m-n)\bmod 1000 = 1000\bmod 1000 = \boxed{000}$.

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

Perform a similar operation as in Solution 2, but only on $y$: $4x+3y+2z=2009$ if and only if $4x+3(y-3)+2z=2000$. The value $m-n$ is thus the additional solutions we have for this equation as opposed to the second.

The second equation $4x+3y+2z=2000$ requires $y$ to be even, so the largest solution for $y$ would be $3y=1992$ and the smallest $3y=6$. Now let $y'=y-3$. Note that we can make $3y'=3(y-3)=1992$ without problems, and likewise with all the solutions in between; so the only additional solutions for $3y'$ will be below the smallest for $3y$. Accordingly, $3y'$ can equal $0$ or $-6$ when $y=3$ or $y=1$.

So there are only two equations to consider: $4x+0+2z=2000$ and $4x-6+2z=2000$, or $2x+z=1000$ and $2x+z=1003$.

If $2x+z=1000$, $x$ must be between $1$ and $499$ to obtain positive integer solutions because $z=0$ when $x=500$; there is exactly one valid $z$ for each $x$.

If $2x+z=1003$, $x$ must be between $1$ and $501$ to obtain positive integer solutions; there is exactly one valid $z$ for each $x$.

Therefore $m-n = 499 + 501 = 1000; 1000 \bmod 1000 = \boxed{000}$.

See Also

2009 AIME II (ProblemsAnswer KeyResources)
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
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