Integers

[willpower]

Let m and n be positive integers such that 7/10 < m/n < 11/15. Find the smallest possible value of n.

[nafistiham]

\[\frac{7} {10}=\frac {42} {60}\]

\[\frac{11} {15}=\frac {44} {60}\]

s0,

\[\frac{m} {n}=\frac{43} {60}\]

\[n=60\]

সমাধানটি ভুল।

[willpower]

Thank you!

[sourav das]

What?? There are infinitely many fractions between $\frac{42}{60}$ and $\frac{44}{60}$.
Just put $\frac{m}{n}=\frac{5}{7}$
My proof:

$\frac{7}{10}< \frac{m}{n}\Rightarrow 7n<10m $ ....(i)
and
$\frac{m}{n}< \frac{11}{15}\Rightarrow 15m<11n$....(ii)
Now multiply (i) with 3 and (ii) with 2 and with together :
$21n<30m<22n$
Now just plug in n = 1,2,3...7;
and see that 7 works.

[willpower]

Another question on integers:
- Prove that the product of four consecutive positive integers is never a perfect square.

[Tahmid Hasan]

Another question on integers:
- Prove that the product of four consecutive positive integers is never a perfect square.

well that's pretty simple,try to prove the product of four consecutive integers is $1$ less from a perfect square.
you can also use Erdos' theorem on the product of consecutive integers.