## Pages

### RMO,INMO PREVIOUS YEAR SOLVED PAPER-OLYMPIAD PRIMER

Here is some solved problems from previous years questions of INDIAN NATIONAL MATHEMATICAL OLYMPIAD & REGIONAL MATHEMATICAL OLYMPIAD.

CHAPTER-INEQUALITIES
Q.[INMO-1991] Let a,b,c be real numbers with 0<a<1,0<b<1,0<c<1 and a+b+c=2. Prove that
a/(1-a) . b/(1-b) . c/(1-c) ≥ 8.

Sol:- Putting x=1-a, y=1-b & z=1-c.
Subject to the condition 0<x,y,z<1, x+y+z=1.
Now, a/(1-a) . b/(1-b) . c/(1-c)
= [(y+z)(z+x)(x+y)]/xyz
={(y+z)/√yz}{(x+y)/√xy}{(x+z)/√xz}
≥ 8, by applying AM≥GM in each element of the above terms.
Hence the proof.

CHAPTER-FUNCTIONS
Q.[INMO-1992] Determine all functions f satisfying the functional relation
f(x) + f(1/(1-x)) ={2(1-2x)}/{x(1-x)} ,for x ≠ 0,1.

Sol:- f(x) + f(1/(1-x)) ={2(1-2x)}/{x(1-x)}
=2/x - 2/(1-x) .
Replace x by 1/(1-x) ,we get
f(1/(1-x))+f(1-1/x)=-2x + 2/x .---(1)
Replace x by 1-1/x ,we get
f(1-1/x) + f(x) =2x/(x-1) - 2x.----(2)
(2)-(1) gives
f(x) - f(1-1/x) =2x - 2x/(1-x).----(3)
Adding 2) & (3) we get
f(x) = (x+1)/(x-1) .

CHAPTER-COMBINATORICS
Q.[RMO-2000] All the 7-digit numbers containing each of the digits 1,2,3,4,5,6,7 exactly once & not divisible by 5, are arranged in the increasing order. Find the 2000th number in this list.
Sol:- The number of 7-digit numbers with 1 in the left most place & containing each of the digits 1,2,3,4,5,6,7 exactly once is 6!=720. But 120 of these end in 5 & hence are divisible by 5. Thus,the number of 7-digit numbers with 1 in the left most place & containing each of the digits 1,2,3,4,5,6,7 exactly once but not divisible by 5 is 600.
Similarly the number of 7-digit numbers with 2 & 3 in the left most place & containing each of the digits 1,2,3,4,5,6,7 exactly once but not divisible by 5 is also 600 each. These amount for 1800 numbers. Hence,2000th number must have 4 in the left most place.
Again the number of such 7-digit numbers beginning with 41,42 and not divisible by 5 is 120-24=96 each & these amount for 192 numbers. This shows that 2000th number in the list must begin with 43.
Thus,the 2000th number is 4315672.

CHAPTER-NUMBER THEORY
Q.[RMO-2006] Find the list possible value of a+b,where a,b are positive integers such that 11 divides a+13b and 13 divides a+11b.
Sol:- 11 | (a+13b) => 11 | (a+2b)
=> 11 | (6a+12b) => 11 | (6a+b).
Since GCD(11,13)=1, we conclude that 143 | (6a+b).
This,we may write 6a+b=143k for all k€N.Hence
6a+6b= 143k+5b=144k+6b-(k+b).
This shows that 6 divides k+b & hence k+b≥6. We therefore obtain
6(a+b)=143k+5b=138k+5(k+b)
≥ 138 + 5*6 =168.
Then a+b≥28.
Taking a=23,b=5 we see that the conditions of the problem are satisfied. Thus the and is 28.

CHAPTER-THEORY OF EQUATIONS
Q.[INMO-2000] If a,b,c,x are real numbers such that abc ≠ 0 and
{xb+(1-x)c}/a ={xc+(1-x)a}/b ={xa+(1-x)b}/c ,
then prove that either a+b+c=0 or a=b=c.

Sol:-{ xb+(1-x)c}/a ={xc+(1-x)a}/b ={xa+(1-x)b}/c ={xb+(1-x)c+xc+(1-x)a+xa+(1-x)b}/(a+b+c) =1.
We get two equations
-a+xb+(1-x)c=0,
(1-x)a-b+xc=0
Solving these two equations, we have a=b=c, since 1-x-x^2 ≠0.