## Shares

From the 2004 Tournament of Towns:

Each day, the price of the shares of the corporation “Soap Bubble, Limited” either increases or decreases by n%, where n is an integer such that 0<n<100. The price is calculated with unlimited precision. Does there exist an n for which the price can take the same value twice?

The Russians are at it again with their well conceived, intriguing maths problems.

What is really being asked is can we find a solution to

$(1+\frac{n}{100})^a(1-\frac{n}{100})^b=1$ ?

$(100+n)^a(100-n)^b=100^{a+b}$.

Now the right hand side is divisible by the primes 2 and 5 only. This severely restricts the possible values of $n$; in fact, the only value of $n$ in the given range that is even worth trying is $n=60$. That is,

$160^a\times 40^b=100^{a+b}$.

Alas, this fails: 2 appears to a higher power than 5 in both 160 and 40, but 2 and 5 must appear equally often on the right hand side. Thus there is no value of n for which the price can take the same value twice.

It is another fantastic problem from that part of the world that produces more interesting problems than anywhere else. I just wish there had been a solution – then the problem would have been that much sweeter…