11.5. For which natural numbers $n$ do there exist natural numbers $a$ and $b$ such that $n!=2^{a}+2^{b}$?
Answer: for $n=3$ or $n=4$.

Solution. Consider the remainders of the powers of two when divided by 7: 1, 2, and 4. This means that the sum of two powers of two cannot give a remainder of 0 when divided by 7, so the left side of the equation is not divisible by 7. Therefore, $n!$ is not divisible by 7, which means $n<7$. We also have $5!=120>2^6+2^5$ and $2^{10}=2^9+2^9>2^9+2^8>720=6!>2^8+2^8$.

Grading Criteria. 
“+” A complete and justified solution is provided 
“士” The solution is logically reduced to a finite enumeration, but it is not performed or is performed with an error 
“干” Only the correct answer with the corresponding values of $a$ and $b$ is provided 
“-” The problem is not solved or is solved incorrectly