# Thread: Need Help With Basic Prob

1. ## Need Help With Basic Prob

que1> Find Number Of Zero's In The Expansion Of 79!?

que2>find Last Two Digit Of 95!?

do Snd The Method Also To Solve These Prob ?

2. Hello, speedtijo!

1) Find the number of zero's in the expansion of $\displaystyle 79!$

Consider the number: .$\displaystyle 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdots 77\cdot78\cdot79$

The question becomes: how many 5's are in the factorization?
(Since "5 times an even number" produces a 0.)

Every fifth number has a 5: .$\displaystyle \left[\frac{79}{5}\right] \:=\:15$
. . There are fifteen 5's among the factors.

But a few of them two 5's . . . these are the multiples fo 25.
. . There are: .$\displaystyle \left[\frac{79}{25}\right] \:=\:3$ of them.
Each of these contribute an addition 5 to the factoring.

Hence, there are: .$\displaystyle 15 + 3 \:=\:{\bf18}$ factors of 5.
. . Each of them, paired with a factor 2, produces a 10.

Therefore, $\displaystyle 79!$ has eighteen 0's at the end.

2) Find the last two digits of $\displaystyle 95!$
As we have seen, 79! ends in 18 zeros.

Multiplying by 80, 81, 82, 83, ... 95, we get even more zeros at the end.

Obviously, the last two digits are 00.