1. A professor from the Department of Mathematical Modeling at FEFU last academic year gave 6480 twos, thereby overachieving the commitments taken at the beginning of the year. In the next academic year, the number of professors increased by 3, and each of them started giving more twos. As a result, a new record was set for enclosed spaces: 11200 twos in a year. How many professors were there initially, if each professor gives the same number of twos as the others during the session?
1. The professors of the Department of Mathematical Modeling at DVFU in the previous academic year gave 6480 twos, thereby over-fulfilling the obligations they had taken on at the beginning of the year. In the next academic year, the number of professors increased by 3, and each of them started giving more twos. As a result, a new record for enclosed spaces was set: 11200 twos in a year. How many professors were there initially, if each professor gives the same number of twos in a session as the others?

Let the number of professors be $p$. Then the number of twos given by each of them is $6480 / p$. This number is less than the number of twos given in the next academic year $-11200 /(p+3)$. Therefore, $p \geq 5$. Each professor gives an integer number of twos, so $6480=2^{4} 3^{4} 5$ is divisible by $p$ and $11200=2^{6} 5^{2} 7$ is divisible by $p+3$. The number $p$ cannot be divisible by 3, since 11200 is not divisible by 3. Therefore, $p=5^{m} 2^{n} \geq 5$, where either $m=0 ; n=3,4$, or $m=1 ; n=0,1,2,3,4$. In the first case, we get values $p+3=11$ or $p+3=19$, which are not divisors of 11200. If $m=1$, then $p+3=13,23,43,83$ for $n=1,2,3,4$ and therefore the only option that satisfies the conditions is $m=1, n=0, p=5$.