please show me the right method

no offense but are you sure you got the question correct??

agreed.. the question, as asked, is impossible.

six digits no multiplied by 9 resulting into another 6 digit no.... weird

Its nt 6 digits......its 2^x3^y5^z...

i think the question states correctly cuz I run into an exactly same question and having problem with....
can anyone help...the answer should be 20.

How can 2xxxxx *9 = 2xxxxx??

Something was mixed up in the question... it's probably something subtle.. but i'm fairly certain it's there.

raunekk wrote:Its nt 6 digits......its 2^x3^y5^z...

Ahh there you go... In that case it's fairly easy.

let's assume that V = 111

So we get 2^1*3^1*5^1 = 2*3*5 = 30

f(V) = 30
Therefore F(M) = 270

If we factor 270, we find 3*3*3*5*2 which is equal to 2^1*3^3*5^1

Therefore M = 131

131-111 = 20

Or alternatively you could just give it some thought. Since we know f(m) has to be 9 times bigger than f(v), and 9 = 3^2.. we know that there will always be two more 3s in the factor of f(m) than f(v). Which leads you to the answer of 20.

thanks egybs...i made an assumption tat x,y, and z cant be equal.. but as such i tried wit x=1,y=2 n z=3.. and got d answer... thanks:)




btw can u give me some advice as to how to improve my score in quant.. i m a bit weak at it..i dont seem 2 develop tat logic.!!can u suggest me some good books.. i am weak at permutation combination,number system and word problems...pls advice