CAT problems

Yesterday i gave CAT 2007. I had no intentions of doing MBA at this stage of my life, but for the satisfaction of my father, i decided to attempt CAT. My preparations amounted to 4 Quants problems and 4 DI problems... :-)...So there was nothing to worry about..i knew i was going to cup, even if put up the best show of my life till now! ... so i took a decision...i will enjoy these two and half hours of my life...and since i was interested in maths, i decided i will do only quants... :-)...and DI if i have time to do so for i am very slow AND have this unique characteristic to commit ridiculous mistakes, no i would rather say, mistakes which are superhuman in nature...that cannot be committed by any normal human !....So with this "strategy" i decided to "attack" the CAT 2007 :-P...Though overall i found CAT very very boring compared to JEE screening or mains(obba)...there were some interesting questions in quants section... i thought it would be good to post solutions to them .

Of all I liked this one particularly. Find all the positive integer pairs (m,n) satisfying the equation 1/m + 4/n = 1/12 , where n is an odd integer less than 60.
This question is very similar to a question which i appeared in INMO many years ago, i believe. Our general tendency to solve this problem is to express one variable in terms of other and using the fact that both are "positive integers", look for the feasible solutions.

But there is another interesting substitution that solves such problems very quickly. One can clearly see that m>= 12 and n>=48. So substitute m=12+k and n=48+l. Now substituting and proceeding some 2 steps(canceling, rearrangements etc.) ,it reduces to k.l = 12x48=(2^6)x(3^2). Since n is odd, l is also odd and the only possible integral values of l are 1,3,9 giving n=49,51,57 all of which are odd integers. So there are in all 3 possible solutions to this equation.

The special feature of this substitution is that it provides exact solutions as long as m,n remain integers(preferably natural numbers, to avoid painful +ve and -ve possibilities in case of domain of Integers !)...I was happy to solve this problem, but unfortunately, my innate ability to commit mistakes made its appearance and instead of adding values of l to 48, i was subtracting it and cursing the IIM's for giving wrong question ! (m becomes -ve then :-P).

Another problem which i found interesting was the 4-digit square problem.This the problem statement Consider 4-digit numbers for which the first 2 digits are equal and the last 2 digits are also equal. How many such numbers are perfect squares?
This problem again has made its appearance in RMO'91(India). Here is a link to the solution to this problem. It uses the divisibility by 11 property of the number.

Since I had seen this prob earlier, i had vague remembrance of this number as 7766, soon i figured out it was 7744, but couldn't prove there were others....But the above solution which uses divisibility by 11 property is quite elegant. All that was required to crack this prob was to observe the above fact ! and all those who did it, cracked it in less than 2 minutes, i believe.

These were the only 2 problems that i found interesting... btw that doesn't mean i solved all other problems... i didn't understand 2 of them and committed several mistakes in others :-)... So in all i cupped in quants too :( ... well that's me :-D ... hope this does some good to those who want to know the solutions :-).

The Inception

Finally...I have a parchment before me, pen in my hand and the mind in its place...All that is left is to 'pen down the adventures of the mind'....and thus it begins...