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### 3 SOLVED PROBLEMS FROM ISI,CMI,RMO PAPER

#### 3 FAQs IN MATHEMATICS

This is for those who are preparing for OLYMPIAD,ISI Entrances,CMI entrances & also for all MATHEMATICS LOVERS.
PROBLEM.1. Show that A=10101 . . . .101 is not a prime number,unless A=101.

Sol:-A= 10^{2n} +10^{2n-2} + . . . . +10^{2} +1 ----------(1)
100A=10^{2n+2} +10^{2n} + . . . +10^{4} +10^{2} ---------(2)
(2)-(1) gives
99A=10^{2n+2} -1 =(10^{n+1} +1)(10^{n+1} -1)
When n>1, 10^{n+1} -1>99. If n is odd,then
10^{n+1} -1 =9999 or 999999, or . . . and the second factor is divisible by 99,except when n=1.
The quotient is greater than 1 and A has two proper factors. If n is even
10^{n+1} +1=1001,100001, . . . and all these are divisible by 11.
So, A is non-prime(or,composite) except when it is 101.

PROBLEM.2.If n be a positive integer & define f(n)=1!+2!+3!+ . . . +n!.Find the polynomials P(x) and Q(x) such that
f(n+2)= P(n)f(n+1) + Q(n)f(n),for all n≥1.

Sol:- f(n)=1!+2!+3!+ . . . +n!
& f(n+1)=f(n)=1!+2!+3!+ . . . +n!+(n+1)! .
So, f(n+1)-f(n)=(n+1)!
Also, f(n+2)=f(n+1)+(n+2)!
=f(n+1)+(n+2)(n+1)!
=f(n+1)+(n+2){f(n+1)-f(n)}
=(n+3)f(n+1) - (n+2)f(n);
So,we have P(n)=n+3 ;  Q(n)=-(n+2).
Hence, P(x)=x+3 & Q(x)=-x-2.

PROBLEM.3.Show that 1< 1/1001 +1/1002 + . . . . +1/3001 <4/3.

Sol:-  A.M. of 2001 numbers,i.e., 1001,1002, . . . . ,3001 =1/2001 {1001+1002+ . . . +3001}=2001/2 *(1001+3001)*1/2001 =2001.
H.M.=2001/{1/1001+1/1002+ . . . . +1/3001}
Since AM>HM
So, 1/1001 +1/1002 + . . . . +1/3001 > 1 so one-half of the inequality is proved.
1/1001 +1/1002 + . . . . +1/1250 <250/1000=1/4;
1/1251 +1/1252 + . . . +1/1500< 250/1250=1/5 ;
1/1501 +1/1502 + . . . +1/2000 <250/1500=1/3 ;
1/2001 +1/2002 + . . . +1/3001 < 1001/2000.
Adding throughout
1/1001 +1/1002 + . . . . +1/3001 < 1/4 +1/5 +1/3 +1001/2000 < 7703/6000 < 4/3 < 3/2.