IMO Shortlist 2011 problem N1
Dodao/la:
arhiva23. lipnja 2013. For any integer
![d > 0,](/media/m/3/1/2/3129ad17dcdf5b139607b4c80aa87a37.png)
let
![f(d)](/media/m/0/a/c/0acf5c8815514967da622b421b7e07df.png)
be the smallest possible integer that has exactly
![d](/media/m/f/7/d/f7d3dcc684965febe6006946a72e0cd3.png)
positive divisors (so for example we have
![f(1)=1, f(5)=16,](/media/m/e/c/4/ec46d07ae930e69d7940cc8ced262fbc.png)
and
![f(6)=12](/media/m/a/6/9/a6986570366ed3f4a79a0186813648b3.png)
). Prove that for every integer
![k \geq 0](/media/m/7/3/a/73a323fe1d8883d41d88552074e84edb.png)
the number
![f\left(2^k\right)](/media/m/4/d/7/4d72e6e875a268a1f3fbdef055fd8490.png)
divides
![f\left(2^{k+1}\right).](/media/m/d/0/3/d03a97845050ba78e9a39035c9998e4d.png)
Proposed by Suhaimi Ramly, Malaysia
%V0
For any integer $d > 0,$ let $f(d)$ be the smallest possible integer that has exactly $d$ positive divisors (so for example we have $f(1)=1, f(5)=16,$ and $f(6)=12$). Prove that for every integer $k \geq 0$ the number $f\left(2^k\right)$ divides $f\left(2^{k+1}\right).$
Proposed by Suhaimi Ramly, Malaysia
Izvor: Međunarodna matematička olimpijada, shortlist 2011