Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$2x + 3y = n$ has exactly $2011$ non-negative integral solutions. Determine the SUM of the possible values of $n$.

share|improve this question
2  
By "integral solutions", I assume you mean pairs $(x,y)\in(\Bbb Z^{\geq 0})^2$? I initially thought that you meant triples $(x,y,n)\in(\Bbb Z^{\geq 0})^3$, in which case your statement would be patently false.... –  Cameron Buie Dec 20 '12 at 4:51

3 Answers 3

up vote 2 down vote accepted

The integer solutions to $2x+3y=1$ are given by

$$\left\{\begin{align*} &x=-1+3k\\ &y=1-2k \end{align*}\right.$$

for $k\in\Bbb Z$, so the integer solutions to $2x+3y=n$ are given by

$$\left\{\begin{align*} &x=-n+3k\\ &y=n-2k \end{align*}\right.$$

for $k\in\Bbb Z$. Clearly $n>0$, so $x\ge 0$ iff $3k\ge n$ iff $k\ge\frac{n}3$, and $y\ge 0$ iff $2k\le n$ iff $k\le\frac{n}2$. Thus, for a given $n>0$ you get one solution to $2x+3y=n$ in non-negative integers for each integer $k$ satisfying $$\frac{n}3\le k\le\frac{n}2\;.\tag{1}$$

$(1)$ is equivalent to $$\left\lceil\frac{n}3\right\rceil\le k\le\left\lfloor\frac{n}2\right\rfloor\;,$$

so there are $\left\lfloor\frac{n}2\right\rfloor-\left\lceil\frac{n}3\right\rceil+1$ solutions. Thus, you’re looking for the sum of all positive integers $n$ such that

$$\left\lfloor\frac{n}2\right\rfloor-\left\lceil\frac{n}3\right\rceil+1=2011$$ or, equivalently,

$$f(n)\triangleq\left\lfloor\frac{n}2\right\rfloor-\left\lceil\frac{n}3\right\rceil=2010\;.$$

Clearly $$f(n)\approx\frac{n}2-\frac{n}3=\frac{n}6\;,$$ so the desired values of $n$ will be near $6\cdot2010=12060$. One of these values is $12060$, since $f(12060)=6030-4020=2010$. Now note that

$$\begin{align*} f(k+6)&=\left\lfloor\frac{k+6}2\right\rfloor-\left\lceil\frac{k+6}3\right\rceil\\ &=\left(\left\lfloor\frac{k}2\right\rfloor+3\right)-\left(\left\lceil\frac{k}3\right\rceil+2\right)\\ &=\left\lfloor\frac{k}2\right\rfloor-\left\lceil\frac{k}3\right\rceil+1\\ &=f(k)+1 \end{align*}$$

for all $k$, so each arithmetic progression with constant difference $6$ contains exactly one term $n$ such that $f(n)=2010$, and there are therefore exactly $6$ such values of $n$. From here you should have no trouble finding the $6$ values of $n$ such that $f(n)=2010$ and their sum.

share|improve this answer

Hint: if $n$ is odd, one solution will have $y=1, x=\frac {n-3}2$ Others will differ by adding $2$ to $y$ and deducting $3$ from $x$ a number of times-what is the maximum number of times allowed? If $n$ is even, what is the base solution?

share|improve this answer

Let us verify for $n=m[2,3]+k$ where $0\le k<[2,3]$

If $k=0\implies 2x+3y=6m\implies 3(2m-y)=2x,$ so $y$ must even $=2z$(say), So, $x=3(m-z)\implies 0\le z\le m$ so $z$(hence $y$) can assume $m+1$ values in non-negative integers.

If $k=1\implies 2x+3y=6m+1\implies 3(y-2m)+1=2x,$ so $y$ must odd $=2z+1$(say), where $z\ge 0$ So, $x+1=3(m-z)\implies (m-z)\ge 1\implies 0\le z\le m-1$ so $z$(hence $y$) can assume $m$ values in non-negative integers. $\implies n=6(m+1)+1=6m+7$ will have $m+1$ solutions in non-negative integers.

So, we observe that $6m,6m+2,6m+3,6m+4,6m+5,6m+7$ will have $m+1$ solutions in non-negative integers.

share|improve this answer
    
A big hint for what looks like homework, but well done. +1 –  Ross Millikan Dec 20 '12 at 5:42
    
This doesn’t work: for $n=2m=4020$, for example, there are only $671$ solutions, not $2011$. –  Brian M. Scott Dec 20 '12 at 6:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.