Problem Statement
Does the equation\[2^m=a_1!+\cdots+a_k!\]with $a_1<a_2<\cdots <a_k$ have only finitely many solutions?
Categories:
Number Theory Factorials
Progress
Asked by Burr and Erdős. Frankl and Lin [Li76] independently showed that the answer is yes, and the largest solution is\[2^7=2!+3!+5!.\]In fact Lin showed that the largest power of $2$ which can divide a sum of distinct factorials containing $2$ is $2^{254}$, and that there are only 5 solutions to $3^m=a_1!+\cdots+a_k!$ (when $m=0,1,2,3,6$).See also [404].
Source: erdosproblems.com/403 | Last verified: January 14, 2026