Combinatoric

Question

Find number of zero's at the end of (2018)! ?

Answer 1

#(zeros)= # pairs of 5's and 2's in 2018!

# 5's in 2018! = $\left \lfloor 2018/5 \right \rfloor + \left \lfloor 2018/25 \right \rfloor + \left \lfloor 2018/125 \right \rfloor + \left \lfloor 2018/625 \right \rfloor$ =502

Obviously # 2's > # 5's.

So, # pairs(No of trailing zeros) = 502 which is min(#5s,#2s)

Comments

Yes, 502 is correct.

---

@shriram s 1 why we are doing the min of 2's and 5's?

I think it is not striking to my mind.

---

@shaktisingh

$10!=10\times 9\times8\times7\times6\times5\times4\times3 \times  2\times 1=10\times 5 \times 2 = 10\times 10 = 100$,so number of 0's at the end is $2.$

Procedure: $10=2\times 5$

Number of $2's$ in $10! = \left \lfloor \dfrac{10}{2}\right \rfloor=5$

$\left \lfloor \dfrac{5}{2}\right \rfloor=2$

$\left \lfloor \dfrac{2}{2}\right \rfloor=1$

So, number of $2's$ is  $= 5+2+1=8$

And 

Number of $5's$ in $10! = \left \lfloor \dfrac{10}{5}\right \rfloor=2$

So, number of $5's$ is  $= 2$

We can write like this $10!=2^{8}\times 5^{2}=2^{6}\times 2^{2}\times5^{2} = 2^{6}\times10^{2}$

So, number of $0's$ at the end is $2.$

Answer 2

The number of $0's$ at the end is equal to the number of $5's$ at the end(Because of $10=2\times5$ and in any $N!\:\:\:\#5's<\#2's$).

$\left \lfloor \dfrac{2018}{5}\right \rfloor=403$

$\left \lfloor \dfrac{403}{5}\right \rfloor=80$

$\left \lfloor \dfrac{80}{5}\right \rfloor=16$

$\left \lfloor \dfrac{16}{5}\right \rfloor=3$

So, total number of $0's$ at the end $=403+80+16+3=502$