Divisors of factorials III

In general, the largest power $m$ of a prime $p$ that divides $n!$ is given by $\displaystyle m = \sum_{k=1}^\infty \left \lfloor \frac n{p^k} \right \rfloor$ . From $n! = (2^a)(3^b)(5^c)(7^d)(11^2)(13^2)(17)(19)(23)$ . the factor $13^2$ tells us that $n \ge 26$ . Since the prime factor 29 is not present, $n < 29$ . That is $n$ can only be 26, 27, and 28. Then we have:

$\begin{aligned} a_{26} + b_{26} + c_{26} + d_{26} & = \sum_{k=1}^\infty \left \lfloor \frac {26}{2^k} \right \rfloor + \sum_{k=1}^\infty \left \lfloor \frac {26}{3^k} \right \rfloor + \sum_{k=1}^\infty \left \lfloor \frac {26}{5^k} \right \rfloor + \sum_{k=1}^\infty \left \lfloor \frac {26}{7^k} \right \rfloor = 23 + 10 + 6 + 3 = 42 \\ a_{27} + b_{27} + c_{27} + d_{27} & = \sum_{k=1}^\infty \left \lfloor \frac {27}{2^k} \right \rfloor + \sum_{k=1}^\infty \left \lfloor \frac {27}{3^k} \right \rfloor + \sum_{k=1}^\infty \left \lfloor \frac {27}{5^k} \right \rfloor + \sum_{k=1}^\infty \left \lfloor \frac {27}{7^k} \right \rfloor = 23 + 13 + 6 + 3 = \boxed{45} \\ a_{28} + b_{28} + c_{28} + d_{28} & = \sum_{k=1}^\infty \left \lfloor \frac {28}{2^k} \right \rfloor + \sum_{k=1}^\infty \left \lfloor \frac {28}{3^k} \right \rfloor + \sum_{k=1}^\infty \left \lfloor \frac {28}{5^k} \right \rfloor + \sum_{k=1}^\infty \left \lfloor \frac {28}{7^k} \right \rfloor = 25 + 13 + 6 + 4 = 48 \end{aligned}$

Therefore, there is only $\boxed 1$ $n$ that satisfies the conditions.

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Notation: $\lfloor \cdot \rfloor$ denotes the floor function .