Problem:
Let b≥2 be an integer. Call a positive integer nb-eautiful if it has exactly two digits when expressed in base b, and these two digits sum to n. For example, 81 is 13-eautiful because 81=6313 and 6+3=81. Find the least integer b≥2 for which there are more than ten b-eautiful integers.
Solution:
Let uvb be a b-eautiful integer. Then 1≤u≤b−1,0≤v≤b−1, and
u+v=sandub+v=s2
for some positive integer s.
Solving this system for u and v in terms of s and b gives
u=b−1s(s−1)andv=b−1s(b−s).
Thus 1<s<b and s(s−1) must be a multiple of b−1. Conversely, if s satisfies these conditions, then u=b−1s(s−1) is an integer satisfying 1≤u≤s−1 and thus v=s−u is a positive integer less than b. Hence for a given base b, the number of b-eautiful integers is equal to the number of positive integers s such that 1<s<b and b−1∣s(s−1).
Let the prime factorization of b−1 be p1e1p2e2…pkek, where k is the number of distinct prime numbers in the factorization. Then
s≡0 or 1(modpiei)for all i∈{1,2,…,k}.
By the Chinese Remainder Theorem, there are exactly 2k solutions for s in the interval [1,b−1]. However, one of these solutions is s=1, which does not lead to valid u and v. Hence the number of b-eautiful integers is equal to 2k−1. The required condition on b is equivalent to
2k−1≥10, implying that k≥4.
The least possible value of b−1 is then 2⋅3⋅5⋅7=210, so the least possible value of b is 210+1=211.