Another fact on factorial

Number Theory Level 4

Let $T(x)$ denote the number of trailing zeros in $\prod_{n=1}^{x} n!$ for a positive integer $x$ .

Let $\{x_1,x_2,\cdots,x_p\}$ be the set of all positive integers $x_i$ which satisfy the equation

$T(x_i)-x_i+1=0$

Find $T\left(\sum_{i=1}^p x_i\right)-\sum_{i=1}^p x_i+1$

The answer is 1.

1 solution

Janardhanan Sivaramakrishnan
Jul 10, 2015

It is easy to see that $x=1$ is a solution to the equation.

Further, it can also be seen that $T(x)$ is a monotonically increasing function for $x \ge 5$ , and has a slope greater than one for $x \ge 10$ , whereas $x-1$ always has a slope of one.

This means that $T(x)-x+1$ decreases for $1 \le x \le 4$ , is constant for $5 \le x \le 9$ and is increasing for higher values of $x$ .

This would further mean that there is no solution in the range $1 < x \le 9$ (as $T(x)<x-1$ in the range)

On search in $x \ge 9$ range, we can see that $x_2=12$ is another solution.

Now, since the slope of $T(x)$ is greater than the slope of $x-1$ for $x>12$ , it can be said that there would be no further solutions.

Hence, the required answer is

$T(1+12)-(1+12)+1=T(13)-12=\boxed{1}$

Another fact on factorial

The answer is 1.

1 solution

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