gmat sets

[bug]

The function f is defined for each positive three-digit integer n by f(n) = 2x3y5z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m)=9f(v), them m-v=?

(A) 8

(B) 9

© 18

(D) 20

 

sorry abt the confusion: f(n) = 2(raised to the power x)3(raised to the power y)5(raised to the power z)

hope this helps

[panchamuga]

Hello mr.bug,

seems to be a great qn...2x3y5z Is it a number???

If u proceed like assuming m=100x+10y+z ,n=100a+10b+c.

Looks like it will not be that easily solvable..

May be i am wrong..

So I tried a diff approach.

looks like f(x) is a 6 digit number.Multiplying it by 9 yields a 7 digit number even if x,y,z are zero...Then howss it possible to have f(m)=9f(n)..

Doubtfull

Help reqd MR.bug...

[chatru]

I am not able to solve:I hope the question is correctly stated

[bug]

manish, gmathelp, any thoughts?

[manish8109]

Is Ans (D) 20 as,

 

let M= abc & V=xyz

 

=> f(m) = 2^a * 3^b * 5^c

f(v) = 2^x * 3^y * 5^z

 

=> f(m) = 9*f(v)

2^a * 3^b * 5^c = 9 * 2^x * 3^y * 5^z

=> 2^a * 3^b * 5^c = 2^x * 3^(y+2) * 5^z

Comparing both sided

 

a=x , b=y+2 , c=z

so,

Let abc=121 then Xyz= 101

=> 121-101= 20 ----Ans

[GMAT-HELP]

Same approach as Manish..

2^x1*3^y1*5^z1 = 3^2*2^x2*3^y2*5^z2

=> 2^x1*3^y1*5^z1 = 2^x2*3^(y2+2)*5^z2

=> y1 = y2+2

=> as y1 and y2 represent tens digits, m-v = 2*10 = 20 Ans D.

[bug]

thanks manish and gmat-help. this is a simple problem, but for some reason i kept getting 18 !

[venkat8103]

May be silly..

 

How did u get the equation in powers (exponents).

 

From the Question isnt it 2x1*3y1*5z1..?

 

Manish, or some one could pls explain me..?

[mastermind]

why is "f(v) = 2^x * 3^y * 5^z" and not f(m) whereas f(n) is given by 2^x * 3^y * 5^z.

 

sorry, my basics of functions are nil.

 

can anybody explain?

[XiK]

answer is 20...

using the exponential logic

 

assuming V =111

=> f(V)= 2^1 * 3^1 * 5^1

now f(M) = 9( 2^1 * 3^1 * 5^1 )

=> 2^1 * 3^3 * 5^1 (coz 9=3^2)

so M = 131

 

therefore

131 - 111 = 20

viola!!!!

 

i think the best way to solve these questions is always taking hypothetical numbers...it getts pretty smooth ...

[maverick312]

20 is correct

[ashraf Ali]

May be silly..

 

How did u get the equation in powers (exponents).

 

From the Question isnt it 2x1*3y1*5z1..?

 

Manish, or some one could pls explain me..?

 

well i have the same question!

 

if in 2x3y5z the xyz means exponents then shouldn't it write this way

 

2^x 3^y 5^z...........?

plz explain i m not good in fuction!

[manish8109]

Hey, it was a typo error in the question,The correct form of the question is in exponential form.

 

Hope it clear ur doubts.

[ashraf Ali]

thanks manish!

[MikeJung]

good q.

[aashu_79]

The function f is defined for each positive three-digit integer n by f(n) = 2x3y5z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m)=9f(v), them m-v=?

(A) 8

(B) 9

© 18

(D) 20

 

sorry abt the confusion: f(n) = 2(raised to the power x)3(raised to the power y)5(raised to the power z)

hope this helps

 

Let m = 100x + 10y + z, n = 100x' + 10y' + z'

 

f(m)/f(v) = 9

=> 2^(x-x') * 3^(y-y') * 5^(z-z') = 9

 

=> x-x' = 0, y-y' = 2, z-z' = 0

 

m - n = 100(x-x') + 10(y-y') + z-z'

= 20

[hjafferi]

IMO D

 

agree with Manish