**PROBLEM PRIMER FOR MATHEMATICS OLYMPIAD**

*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.

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.

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.

{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.