The function f is defined for each positive three-digit integer n by

Bunuel,

Can you help me with my doubt as mentioned below

I could follow rest, but i could not understand how we interpret this

\(2^a*3^b*5^c=9*2^r*3^s*5^t\) --> \(2^{a-r}*3^{b-s}*5^{c-t}=3^2\) (or \(2^a*3^b*5^c=2^r*3^{s+2}*5^t\)) --> \(a-r=0\), \(b-s=2\) and \(c-t=0\) (or \(a=r\), \(b=s+2\) and \(c=t\))

how do we reach a-r=0 ? because both a and r are on different side with base 2

\(2^a*3^b*5^c=2^r*3^{s+2}*5^t\): exponents of 2, 3 and 5 must be equal --> \(a=r\), \(b=s+2\) and \(c=t\).

Fantastic solution.

How does one go about actually learning this? I sat for 10 minutes not understanding anything. Why is it that f(m) = 3^2f(v) makes v have the same digits but the tenths? I'm completely in the dark

Say,

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

and

\(f(v) = 2^d * 3^e * 5^f\)

Now, you are given that \(f(m) = 3^2 * f(v)\)

So \(2^a * 3^b * 5^c = 3^2 * 2^d * 3^e * 5^f\)
\(2^a * 3^b * 5^c = 2^d * 3^{e+2} * 5^f\)

Now, for left side to be equal to right side,
a = d
b = e+2
c = f

That is, the tens digit of v should be 2 more than the tens digit of m. The units and hundreds digits should be the same.

Follow posting guidelines (link in my signatures). Search for a question before starting a new thread. This question has already been discussed before. Topics merged.

If you want the algebraic solution, refer to Bunuel 's post above the-function-f-is-defined-for-each-positive-three-digit-100847.html#p779584.

For a method with options refer below.

Dont let the definition of the function confuse you.

You are given that m and v are 3 digit positive numbers and the function f(n) is true for ALL positive 3 digit integers. Thus f(n) when n=100 = 2^1*3^0*5^0 = 2 and f(999) = 2^9*3^9*5^9 etc.

Now if lets say v=100, then we have to find that value of m that gives us the relation f(m)=9f(v). Lets use the options given to us to come to the correct answer by noting that m-v=options given ---> m = v+options.

A) 8. m=100+8=108 ---> f(m)=f(108)=2^1*3^0*5^8 and f(v) = f(100)=2^1*3^0*5^0. Clearly f(m)/f(v) \(\neq\) 9. Eliminate this option.

B) 9. m=100+9=109 ---> f(m)=f(109)=2^1*3^0*5^9 and f(v) = f(100)=2^1*3^0*5^0. Clearly f(m)/f(v) \(\neq\) 9. Eliminate this option.

C) 18. m=100+18=118 ---> f(m)=f(118)=2^1*3^1*5^8 and f(v) = f(100)=2^1*3^0*5^0. Clearly f(m)/f(v) \(\neq\) 9. Eliminate this option.

D) 20. m=100+20=120 ---> f(m)=f(120)=2^1*3^2*5^0 and f(v) = f(100)=2^1*3^0*5^0. Now, f(m)/f(v) = 9. This is the correct answer. You can stop here. No need to check the last option.

E) 80. m=100+80=180 ---> f(m)=f(180)=2^1*3^8*5^0 and f(v) = f(100)=2^1*3^0*5^0. Clearly f(m)/f(v) \(\neq\) 9. Eliminate this option.

Hope this helps.

Hi,

1) Everything in this Q is dependent on its meaning..

f(m) = 9f(v)= 3^2f(v)

we know each function is product of 2, 3, and 5 raised to some power..

f(m) = 3^2f(v) tells us the change is only in power of three..
and what is the change --2...

what does the power of 3 represent-- tens digit..
so in m, tens digit increases by 2, or in other words the number is 2*10 more than the other number..

2)The second way would be to convert it into numbers..

m= abc= 100a+10b+c and f(m)= 2^a 3^b 5^c..

v= xyz=100x+10y+z and f(v)= 2^x 3^y 5^z ..

f(m) = 9f(v)= 3^2f(v)..

so 2^a 3^b 5^c=3^2 2^x 3^y 5^z = 2^x 3^(y+2) 5^z ..
equate two sides so a=x, b=y+2, and c=z...

substitute these values in m..
100x + 10(y+2) + z= 100x+10y+20+z..
m-v= 100x+10y+20+z-(100x+10y+z )=20

ans 20
D

The easiest way for me was:
take v=100 as the least possible 3 digits positive integer and plug it into the function
f(v)=2^1*3^0*5^0= 2*1*1=2
Then we know that f(m) = 9*f(v)
so f(m) = 18
it's obvious that we keep 5^0 in order to get 1, then what are prime factors for 18, it's 2 and 3^2
then the 3 digits number is m=120

120-100=20

works like a charm for all that kind of questions

Hi All,

While this prompt is 'complex-looking', it's based on some relatively simple concepts (once you unravel the information that you've been given).

To start, we're told that each DIGIT of a positive 3-digit number is defined by...
f(N) = (2^X)(3^Y)(5^Z) where X, Y and Z are the hundreds, tens and units digits of N.

For example, IF... N = 410, then f(410) = (2^4)(3^1)(5^0) = (16)(3)(1) = 48

We're told that M and V are 3-digit numbers such that f(M) = 9(f(V)). We're asked for the value of M-V.

Given how the function multiplies 'powers of primes', for one 3-digit number to be 9 TIMES another, the larger number MUST have "two more 3s" than the smaller number. For that to occur, the tens digit of the larger number must be 2 greater than the tens digit of the smaller number. The other two digits (hundreds and units) must be the SAME.

For example: 240 and 220
f(240) = (2^2)(3^4)(5^0) = (4)(81)(1) = 324
f(220) = (2^2)(3^2)(5^0) = (4)(9)(1) = 36
324 = 9 times 36

Thus, the difference between the M and V comes down to the tens digits - and while there are several potential options here - they ALL differ by 20.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

The function f is defined for each positive three-digit integer n by \(f(n) = 2^x3^y5^z\) , 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 \(f(m)=9*f(v)\) , then \(m-v=\) ?

Let f(v) = 2^x3^y5^z
f(m) = 9*f(v) = 2^x3^{y+2}5^z

Tenth digit of m is 2 greater than tenth digit of v
m - v = 2*10 = 20

IMO D

The function f is defined for each positive three-digit integer n by \(f(n) = 2^x3^y5^z\) , 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 \(f(m)=9*f(v)\) , then \(m-v=\) ?

The way the question is written makes it appear more intimidating than it actually is.

Given: \(f(n) = 2^x3^y5^z\)

This means if we were given the number 729, it would be written as \(f(n) = 2^73^25^9\)

\(f(m)=9*f(v)\)
\(f(m)=3^2*f(v)\)

\(m-v=\) = \(f(m) = 2^x3^{y+2}5^z\) - \(f(v) = 2^x3^y5^z\)

This seemingly complicated function is simply telling us that the tens digit of m is 2 greater than the tens digit of v.

For example, we have two numbers 143 and 123.

143 - 123 = 20.

Answer is D.

Let ,
\(m=abc\), and

\(v=pqr\)

\(so,\)
\(f(m)= 2^a3^b5^c\)

\(f(v)= 9*2^p3^q5^r\) \(=3^22^p3^q5^r\) \(= 2^p3^{q+2}5^r\)

\(Again,\)
\(f(m)=2^a3^b5^c\)

\(So, \)
\(f(m)= 2^p3^{q+2}5^r\)

Using place value for m
\(100p+10(q+2)+r\)

\(now,\)

\(f(m)-f(v)=100p+10(q+2)+r-100p-10q-r\)

\(= 100p+10q+20+r-100p-10q-r\)

\( =20\)

Ans is 20. D

Hi Experts,

I have a doubt that I can't seem to resolve. When we say the function of x, shouldn't the variable 'x' necessarily be in the equation?

For example f(x) = 5x + 2 etc.

However, in this question f(n) = (2^x) * (3^y) * (5^z) I mean where's 'n'?
f(123) = (2^1) * (3^2) * (5^3) Shouldn't the value 123 itself be in the equation of the function?

Please help!

Hi Vegita,

You bring up a fair point, but your explanation seems to ignore some of the information that prompts provides:

1) N is a 3-digit number
2) X represents the hundreds digit of N, Y represents the tens digit of N and Z represents the units digit of N

Thus, "N" IS in the equation (but it's broken-down into 3 'pieces'). You could rewrite the function as....

f(N) = [2^(hundreds digit of N)] x [3^(tens digit of N)] x [5^(units digit of N)]

GMAT assassins aren't born, they're made,
Rich

760+: What GMAT Assassins Do to Score at the Highest Levels

I went with v = 111 --> f(v) = 30 --> f(m) 30*9 = 270

270 = 3*3*3*10 = 3*3*3*2*5 --> f(m) must be 131

131-111 = 20

Bunuel
Thank you for this. I don't quite follow how you were able to set the equation equal to 3^2. Any advice would be greatly appreciated.

9 is 3^2. I reformatted the solution. Hope now it's clearer.