Welcome to Ctanujit Institute of Statistics & Mathematics...An Initiative by 'ISI'ian



Urgent Notice:

Due to unavoidable reason, Ctanujit Classes will be closed from 1st November to 30th November 2018.
Postal Package will be available from 1st December 2018.

LEARN MATHEMATICS TRICKS ON NUMBER THEORY

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.

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