## 45 Replies to “INMO 2012 Solutions”

1. another solution of the second part:it is simple to see that p1,p2 are twin primes,{p3,p4 }{q1,q2},{q3,q4}we know that all primes are of the form of 6k+1 or 6k-1 (as seen above)hence modulo 6, { p1,p2,p3,p4 } and {q1,q2,q3,q4} are {-1 , 1 ,-1 ,1}now we cannot add any no. that is not cogruent to 0 mod 6 else it cannot give the same set {-1 ,1 ,-1 ,1}hence we conclude that the no. added is a multiple of 6now all primes(>5) modulo 30 are {1,7,11,13,17,19,23,29}out of which only set satisfies that p4-p1=8{11,13,17,19}on similar reasoning we conclude that the no added is 0 modulo 30

2. That is true… we have corrected the misprint… thanks for your contribution… we will love to do mathematical interactions with you. You may also do some guest blogging in our blog.

3. It can be shown that the sets of primes are {p, p+2, p+6, p+8} and {q, q+2, q+6, q+8}. Upon checking modulo 2,3,5 separately, we can conclude that both p and q are congruent to 1 mod2, 2 mod3 and 1 mod 5. Therefore (p-q) is congruent to 0 mod2, 0 mod3 and 0 mod5 i.e. it is divisible by 2,3 and 5. This is the method I used to prove it in the exam.

4. in Q-2, it is easier to see that primes of form 30k+1,30k+7,30k+11,30k+13,30k+17,30k+19,30k+23,30k+29.Only 30k+11,30k+23 can be p1 so that 30k+19 and 30(k+1)+1 will be primes.now between 30k+23 and 30(k+1)+1 only 30k+29 is prime but there has to be 2 primes(p2,p3).but between 30k+11 and 30k+19 there is 30k+13 and 30k+17.so req. problem has p1 only of form 30k+11.let p1=30k+11 q1=30m+11.so p1-q1=30(k-m).hence 30 divides p1-q1.

5. in Q-5,let angle refers to().(ABC) means angleABC.SOLUTION-(AFE)+(AEF)=(AEF)+(CED)=180-(BAC).so (AFE)=(CED).similarly (CED)=(FDB) proceeding with above reasoning.so (AFE)=(CED)=(FDB).likewise proceeding with above reasoning (AEF)=(CDE)=(BFD).now seeing the triangles AFE,CED,BFD we note they are similar applying above results and AA similarity.So (BAC)=(ABC)=(ACB)=180/3=60.Hence triangle ABC is equilateral.{perhaps the question sways us away by providing additional information of angle bisectors,altitudes etc. whereas it is a simple case of angles involved}.

6. Hey, I want to know what is the average cut off marks for this exam. If anyone has idea about it, please send it. And what is cut off for class XII.

7. I'm in class 12 and i gave the inmo exam. I got 3 questions completely correct. Do you think I will clear the exam? I mean only 6 students are selected from the whole country.. So do you think my score is good enough? Please tell me

8. Sir, i have a simpler solution to question-5.We can easily see that triangle ABC and triangle DEF are similar.Also points B,F,C are concyclic with center D, as it is the angle made by a semi-circle. So, BD=DF.Now angle DFE= 2(angle EBD). So,points B,D,E are also concyclic with center F. So, BF=FD=FE which means angle ABC=60 degrees.Now in traingle DEF, FD=FE so angle so angle FDE=angle DEF=angle EFD=angle A= angle B= angle C =60 degrees!So triangle ABC is equilateral. Sir i want to know if i will qualify with doing only this question?!Partha S MishraClass 10

9. Any of one-can u please tell me about cut-off marks for X clas?I have done three of them by using different type of examples.I didnot use any variables.In the last i.e., 6th question I have done half the answer and in that second half I got 10-multiple as the answer. Will I be selected?