Number Theory Problem No.4

The exponent or power $n_p$ of a prime factor $p$ in $N!$ is given by:

$n_p(N) = \sum_{k=1}^{\left \lfloor \log_p N \right \rfloor} \left \lfloor \frac N{p^k} \right \rfloor$

For example,

$\begin{aligned} n_2(1000) & = \left \lfloor \frac {1000}2 \right \rfloor + \left \lfloor \frac {1000}4 \right \rfloor + \left \lfloor \frac {1000}8 \right \rfloor + \left \lfloor \frac {1000}{16} \right \rfloor + \left \lfloor \frac {1000}{32} \right \rfloor + \left \lfloor \frac {1000}{64} \right \rfloor + \left \lfloor \frac {1000}{128} \right \rfloor + \left \lfloor \frac {1000}{256} \right \rfloor + \left \lfloor \frac {1000}{512} \right \rfloor \\ & = 500 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 \\ & = 994 \end{aligned}$

In computation, you can take the integer value of the previous term after divided by $2$ . For example, $\left \lfloor \dfrac {125}2 \right \rfloor = 62$ .

Similarly, $n_3(1000) = 498$ , $n_5(1000) =249$ , $n_7(1000) = 164$ , $n_{11} (1000) = 98$ , and $n_{13}(1000) = 81$ . Therefore $a-b-c+d+e+f = \boxed{590}$ .

To do the computation, I used a simple Excel spreadsheet

By using this formula:

$\displaystyle n!= \prod_{k=1}^{m} a_{ p(k) }^{\displaystyle \sum_{i=1}^{⌊log_{a_{p(k)}n}⌋} ⌊\frac{n}{a_{p(k)}^{i}}⌋}$

with $m$ is the number of prime number $a_{p}$ in domain $[2,n]$ , $a_{p(k)}$ is the $k^{th}$ prime in that domain, and $n$ is the positive integer.

Calculation gives $a-b-c+d+e+f=\boxed{590}$ .