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1990 OIM Problems/Problem 1

Problem

Let $f$ be a function defined in the set of integers greater or equal to zero such that:

(i) If $n=2^j-1$ , for all $n=0, 1, 2, \cdots,$ then $f(n)=0$

(ii) If $n \ne 2^j-1$ , for all $n=0, 1, 2, \cdots,$ then $f(n+1)=f(n)-1$

a. Prove that for all integer $n$ , greater or equal to zero, there exist an integer $k$ grater than zero such that $f(n)+n=2^k-1$

b. Calculate $f(2^{1990})$ .

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Part a.

$f(2^j-1)=0$ , $f(2^j)=-1$ , $f(2^j+1)=-2$ , and so on..

So we pick a range where $f(n)\ne 0$ which is $2^j-1<2^j+b<2^{j+1}-1$ where $b$ is a non-negative integer.

Therefore, $2^j\le2^j+b\le2^{j+1}-2$ which provides the range for $b$ as: $0\le b \le 2^{j+1}-2-2^j$

In that range, this gives $f(2^j+b)=-(b+1)$

When $n=2^j-1$ , $f(n)+n=f(2^j-1)+2^j-1=0+2^j-1=2^k-1$ which proves the equality since $j=k$ .

When $n \ne 2^j-1$ , $f(n)+n=f(2^j+b)+2^j+b=-(b+1)+2^j+b=2^j-1=2^k-1$ which also proves the equality since $j=k$ .

Part b.

$f(2^{1990})=f(2^{1990}-1)-1=0-1=-1$

This seems to be one of the easiest problems I've seen in these types of competitions, unless I'm missing something.

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

https://www.oma.org.ar/enunciados/ibe5.htm

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