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

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.

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.

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.

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}.

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

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

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?

Dear Partha, it is difficult to discuss solutions in the comment section, why don't you join us in the forum http://www.cheenta.com/forum/viewtopic.php?f=2&t=1 post your solution and query there and all members can discuss on it

This comment has been removed by a blog administrator.

i think there is some problem with your logic of second question. how can the last digit of a prime greater than 5 be 2???

@Above. Your point is correct. It cannot be 2, it can only be 7. The rest of the solution is correct.

It is not that last digit of the prime that we are looking at…. it is the last digit of 'k'. For example 11 is a prime of the form 6k – 1 where k = 2.

It is not that last digit of the prime that we are looking at…. it is the last digit of 'k'. For example 11 is a prime of the form 6k – 1 where k = 2.

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

This comment has been removed by a blog administrator.

I also want to add that this is probably the first attempt by anyone at the solutions. Good work.

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.

This comment has been removed by a blog administrator.

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.

This comment has been removed by a blog administrator.

The functional equation's pretty simple, actually…

This comment has been removed by the author.

This comment has been removed by a blog administrator.

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.

@Anon: Recheck your calculations. The quad's area will be less than 2.

This comment has been removed by the author.

@s about 1.8 that is

Hey I got all 6 correct. If you want answers send mail to [email protected]

325 it is!

This comment has been removed by a blog administrator.

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}.

This comment has been removed by a blog administrator.

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.

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

3 questions may take you to IMOTC…. however 4 fully correct attempts is more comfortable…

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

sir you did the second problem in more steps .it can be just done in 9 steps

The cutoff would never be one question in such an easy INMO .

That is very good… please post here your alternate solution..

in class 12 you need to do atleast 5 of them otherwise its not possible until some miracle happens.

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?

WHICH QUESTION NO. IS CHALLENGING ?

I am in 10.I have done 2 and half correctly and in 1 question approach is correct.Whether there are chances for me to attend IMOTC

Sir, is my solution correct?

Dear Partha, it is difficult to discuss solutions in the comment section, why don't you join us in the forum http://www.cheenta.com/forum/viewtopic.php?f=2&t=1 post your solution and query there and all members can discuss on it

please discuss this in the forum… http://www.cheenta.com/forum/viewtopic.php?f=2&t=1

You may join us in the forum for further problems and discussion : http://www.cheenta.com/forum/viewtopic.php?f=2&t=1

You may join us in the forum for further problems and discussion : http://www.cheenta.com/forum/viewtopic.php?f=2&t=1

You may join us in the forum for further problems and discussion : http://www.cheenta.com/forum/viewtopic.php?f=2&t=1

When the result will be declared? For how much time we have to wait approximately?

When the result will be declared? Please tell.

Hey INMO results are out

INMO results are out