#### MATHS-TRICK-2

####
*
This is a kind of initiative from me to publish tricks to state a trick & solve problems using those tricks from tifr,isi,cmi,nbhm papers.*

**Theorem: (de Polinac's formula)**

*Let p be a prime and e be the largest exponent of p such that p^e divides n! ,*

*then e=∑ [n/p^i ]. Where i is running from 1 to infinity.*

**Proof :**n!= 1.2. . . . p. . .2p . . . (p^2). . .(p+1)p . . .(p^3) . . .n ,where [n/p] terms which are divisible by p & among these there are [n/p^2] terms which are divisible by p^2 and so on.

So, the highest power of p that divides n! is

[n/p]+[n/p^2]+[n/p^3]+ . . . +[n/p^k] where p^k is the largest power of p which is less than or equal to n.

Hence we have the proof.

**Problem1. [Olympiad] How many zeros are at the end of 1000! ?**

Sol :- The number of 2's is enough to match each 5 to get a 10 . So,

[1000/5]+[1000/25]+[1000/125]+[1000/625]=249 .

Thus, 1000! ends with 249 zeros.

**Problem 2.[ISI B.STAT 2010] The product of the first 100 positive integers ends with**

**A.21 zeros B.22 zeros C.23 zeros D.24 zeros**

Sol:- (D) Put p=5,n=100,thus from the theorem we have [100/5]+[100/25]=24 zeros as the answer.

**Problem3.[CMI] The highest power of 3 contained in 100! is**

**A.46 B.47 C.48 D.None**

Sol:- (C) Highest power of 3 contained in 100! is

=[100/3]+[100/3^2]+[100/3^3]+ . . . =33+11+3+1+0+0+. . =48.

KEEP VISITING. FOR MORE TRICKS VISIT HERE