2021 USAJMO Problems/Problem 5
A finite set of positive integers has the property that, for each and each positive integer divisor of , there exists a unique element satisfying . (The elements and could be equal.)
Given this information, find all possible values for the number of elements of .
Solution
The answer is (left out by problem author) or for any non-negative integer .
To construct , take distinct primes and construct elements for by taking the product over all indices of either or for every .
Note that (from the problem statement) each element of has divisors.
Now it suffices to show that there is no element such that is divisible by more than one power of any prime.
Suppose there does exist an element and a prime such that .
This implies that there exists an element that is divisible by but not by .
Exactly half of 's divisors are not divisible by , so it follows that exactly half of the elements in are not divisible by .
However, if are the powers of dividing for some , only of the divisors of are not divisible by .
But this means that of the elements of are not divisible by , contradiction.
Therefore, must be some power of .
See Also
2021 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.