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let $a,b$ is positive integer numbers,and such $$\begin{cases} \gcd(a,b)=1\\ b\le 100\\ \left(\dfrac{a}{b}-\left[\dfrac{a}{b}\right]\right)\left[\dfrac{a}{b}\right]=2013 \end{cases}$$ Find the pairs $(a,b)$

where $[x]$ is the largest integer not greater than $x$.

my idea: since $$2013=3\times 11\times 61$$

and let $a=kb+r,0<r<b$,nad we note $$\left[\dfrac{a}{b}\right]\in N,0\le \dfrac{a}{b}-\left[\dfrac{a}{b}\right]<1$$.and then I can't.Thank you

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Let $a=kb+r$, where $k, r \in \mathbb{Z}$ and $0 \leq r<b$. Since $\gcd(a, b)=1$, we have $\gcd(r, b)=1$.

Now the third equation becomes $\frac{r}{b}k=2013$, so $b \mid rk$. Since $\gcd(r,b)=1$, we get $b \mid k$. Let $k=lb, l \in \mathbb{Z}$, so $rl=2013$.

Now $r \mid 2013$ and $r<b \leq 100$, so $r=1, 3, 11, 33, 61$.

At this point, note that each choice of $r, b$ such that $r<b \leq 100, r \mid 2013$ and $\gcd(r, b)=1$ corresponds to exactly one solution for $(a, b)$. (Since $a=kb+r=lb^2+r=\frac{2013}{r}b^2+r$)

If $r=1$, we have $1<b \leq 100, \gcd(b, 1)=1$, giving $99$ solutions for $b$.

If $r=3$, we have $3<b \leq 100, \gcd(b, 3)=1$, giving $97-32=65$ solutions for $b$.

If $r=11$, we have $11<b \leq 100, \gcd(b, 11)=1$, giving $89-8=81$ solutions for $b$.

If $r=33$, we have $33<b \leq 100, \gcd(b, 33)=1$, giving $67-22-6+2=41$ solutions for $b$.

If $r=61$, we have $61<b \leq 100, \gcd(b, 61)=1$, giving $39$ solutions for $b$.

This gives a total of $325$ solutions for $(r,b)$ and hence for $(a, b)$.

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  • $\begingroup$ for the number theory problem,you always give nice solution! Thank you +1 $\endgroup$ – china math Apr 8 '14 at 15:10

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