Question

If n! has 4 zeroes at the end of (n +1)! = n!(n+1) has six zeroes at the end, find the value of n.
(1) 25
(2) 20
(3) 24
(4) 26
​

Answer 1

Concept to be used

• Factorization
• Factorial

1. If we look at the structure of n!

$\sf{n!=1\times2\times3\times...\times{n}}$

It is a multiplication from 1 to n.

2. We don't consider the multiples of 2,

since the multiples of 5 are always less.

3. Since n! has 4 zeros,

we can say n! has 2⁴ × 5⁴ as a factor.

It will always have 4 factors of five.

Solving the Problem

The multiples of 5

• 5, 10, 15, 20, 25, ...

We need the first 4 of the multiples, so that n! has 4 zeros.

$\sf{\therefore{20\leq{n}<25}}$

And using the same method for (n+1)!

• 5, 10, 15, 20, 25, 30, 35, ...

If we look at this closer, 25 has two factors of 5.

$\sf{\therefore{25\leq{n+1}<30}}$

Conclusion

There exist one solution $\sf{n=24}$

$\displaystyle\sf{{\left \{ {{20\leq n<25} \atop {24\leq n<29}} \right. }}$

Therefore the value of n is 24.