1
$\begingroup$

Find the smallest positive integer $m$ such that there exists positive integer $n$ that satisfies that $\lvert {n\over m} - {2\over 5} \rvert\le {1\over100}$

I tried to simplify and turn it into the following $$\lvert {5n-2m\over5m} \rvert\le {1\over100}$$ $$100\cdot\lvert {5n-2m\over5m} \rvert\le1$$ Since $5m>0$ we get that $$100\cdot\lvert5n-2m\rvert\le5m$$ We can take cases now $$\begin{cases} 100(5n-2m)\le5m, & \text { if } & 5n-2m\ge0 \\[2ex] 100(2m-5n)\le5m, & \text { if } & 5n-2m\le0 \end{cases}$$ Unfortunately I am unable to continue from here. Any help would be appreciated.

$\endgroup$
  • 3
    $\begingroup$ What about $m=5$? $\endgroup$ – J. W. Tanner Sep 6 at 11:57
  • 1
    $\begingroup$ Just try each case up to $m=5$ by hand. Try with $1\le n\le m$. $\endgroup$ – Peter Foreman Sep 6 at 12:05
1
$\begingroup$

$m=3:$

if $n\ge2$ then $|\frac n3-\frac25|=\frac n3-\frac25\ge\frac23-\frac25=\frac4{15}\gt\frac1 {100}$

if $n\le1$ then $|\frac n3-\frac25|=\frac25-\frac n3\ge\frac25-\frac13=\frac1{15}\gt\frac1 {100}$

$m=4:$

if $n\ge2$ then $|\frac n4-\frac 25|=\frac n4-\frac25\ge\frac24 - \frac25=\frac1 {10}>\frac1{100}$

if $n\le1$ then $|\frac n4-\frac 25|=\frac 25-\frac n4\ge\frac25-\frac14=\frac3{20}>\frac1{100}$

Cases $m=1$ and $m=2$ are subsumed under case $m=4$, since $\frac n1=\frac {4n}4$ and $\frac n2=\frac {2n}4$.

Therefore $m=5$, where $|\frac25-\frac25|=0<\frac1 {100}$.

$\endgroup$
2
$\begingroup$

$|\frac{n}{m}\,-\,\frac{2}{5}|\,\leq\frac{1}{100} $ $$ |\frac{100n}{m}\,-\,40|\, \leq 1. $$ 39$\leq\frac{100n}{m}\,\leq41$ Therefore $\frac{n}{m}=\frac{2}{5}$ Hence 5n=2m hence min m will be 5 when n=2.

$\endgroup$
  • $\begingroup$ How do you know $\frac nm=\frac25$ from $39\le\frac{100n}m\le41$? There are other fractions $\frac{100n}m$ between $39$ and $41$, e.g., $\frac{900}{22}$. How do you know (as OP asked) that $5$ is the smallest positive integer $m$ satisfying $39\le\frac{100n}m\le41$? $\endgroup$ – J. W. Tanner Sep 6 at 13:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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