thnx frnds and yes ill post in the given thread from now on

35 is the correct solution....

sansdropper SaysQ . n is a number such that 2n has 28 factors and 3n has 30 factors,6n has how many factors?

pls don't start a separate thread for your queries..

you can post quant questions on this thread

http://www.pagalguy.com/discussions/official-quant-thread-for-cat-2011-part-7-25072542

from 1sr statement there are two possibilities...2^2*3^6 or 2^5*3^3

(considering that from 2nd statement on multiplying n with 3 or n with 2 , factors don't double or get hugely increased hence 2 and 3 are the only prime factors)

now from 2nd statement we find that 1st case in not possible hence, our n= 2^5*3^3

now u can find 6n, (6+1)*(4+1) =35

sansdropper Saysyes how did u infer?

its a standard question... if 2n has 28 factors and 3n has 30 factors.... the possible breakup would be 4*7 and 5*6... that means the power of 2 in n is 5 and power of 3 in n is 3.... therefore 6n is (2^6)*(3^4).. thus number of factors is 7*5 = 35

28 = 7 x 4; (for 2N 2^6 x 3^3)

30 = 6 x 5; (for 3N 2^5 x 3^4) for 2N to 3N power of 2 ill reduce by 1 & 3 should increase by 1 , this is the only case possible.

hence N = 2^5 x 3^3 hence 6N = 2^6 x 3^4 & factor = 7x5 = 35

yes how did u infer?

sansdropper SaysQ . n is a number such that 2n has 28 factors and 3n has 30 factors,6n has how many factors?

is answer 35 ?

Q . n is a number such that 2n has 28 factors and 3n has 30 factors,6n has how many factors?