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655-705 Level|   Algebra|   Exponents|   Functions and Custom Characters|   Number Properties|                              
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The function f is defined for each positive three-digit integer n by 11 Sep 2010, 06:40
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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 \(m-v=\) ?

(A) 8
(B) 9
(C) 18
(D) 20
(E) 80
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D
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Bunuel
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The function f is defined for each positive three-digit integer n by 11 Sep 2010, 06:51
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Orange08
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)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80
Show more

Let the 3-digit positive integer \({m}\) be \({abc}\) and positive integer \(v\) be \(rst\). Question: \((100a+10b+c)-(100r+10s+t)=?\).

Then \(f(m)=2^a*3^b*5^c\) and \(f(v)=2^r*3^s*5^t\).

Given:
    \(f(m)=9*f(v)\);

    \(2^a*3^b*5^c=9*2^r*3^s*5^t\);

    \(2^{a-r}*3^{b-s}*5^{c-t}=3^2\) (or \(2^a*3^b*5^c=2^r*3^{s+2}*5^t\)).

So, \(a-r=0\), \(b-s=2\) and \(c-t=0\) (or \(a=r\), \(b=s+2\) and \(c=t\)) --> \(m-v=(100a+10b+c)-(100r+10s+t)=10b-10(b-2)=20\). For example if \(m\) and \(v\) are 143 and 123 respectively (hundreds and units digit are equal and tens digit of \(m\) is 2 more than tens digit of \(v\)) then \(143-123=20\).

Answer: D.

Similar problems:
https://gmatclub.com/forum/k-and-l-are- ... 91004.html
https://gmatclub.com/forum/the-three-di ... 63241.html
https://gmatclub.com/forum/for-any-four ... 26522.html
https://gmatclub.com/forum/k-and-l-are- ... 26646.html

Hope it helps.
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The function f is defined for each positive three-digit integer n by Updated on: 06 Sep 2023, 23:26
Originally posted by KarishmaB on 04 Nov 2010, 18:58.
Last edited by KarishmaB on 06 Sep 2023, 23:26, edited 1 time in total.
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Orange08
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)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80
Show more

The method I would use to solve this problem is exactly the same as saxenashobhit did. I am writing it down just to give a little more detailed explanation.

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

I read the first line and say, "Well, this only means that \(f(146) = 2^1*3^4*5^6\), no matter how sordid it looks."

Now next line tells me that \(f(m)=9f(v) = 3^2 f(v)\). This tells me that m and v are the same except that the ten's digit of m is 2 more than ten's digit of v. This means m - v = 20.
If this is unclear, look at the example above. I can say that if f(v) = \(f(146) = 2^1*3^4*5^6\) then f(m) will be \(3^2 * 2^1*3^4*5^6\) i.e. \(f(m) = 2^1*3^6*5^6\) giving us m as 166. m - v = 166 - 146 = 20
Then m - v = 20

Check this post on functions: https://anaprep.com/algebra-functions/
and this video: https://youtu.be/ik5XijHCOAI
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The function f is defined for each positive three-digit integer n by Updated on: 02 Oct 2022, 23:28
Originally posted by KarishmaB on 08 Mar 2015, 20:41.
Last edited by KarishmaB on 02 Oct 2022, 23:28, edited 1 time in total.
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erikvm
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
Show more

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.
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The function f is defined for each positive three-digit integer n by 04 Nov 2010, 12:40
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Bunuel
Orange08
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)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80

Let the 3-digit positive integer \({m}\) be \({abc}\) and positive integer \(v\) be \(rst\).

Then \(f(m)=2^a*3^b*5^c\) and \(f(v)=2^r*3^s*5^t\).

Given: \(f(m)=9*f(v)\) --> \(2^a*3^b*5^c=9*2^r*3^s*5^t\) --> \(2^{a-r}*3^{b-s}*5^{c-t}=3^2\) (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 \(c=t\)) --> \(m-v=20\), because if for example \(m\) and \(v\) are 143 and 123 respectively (hundreds and units digit are equal and tens digit of \(m\) is 2 more than tens digit of \(v\)) then \(143-123=20\).

Answer: D.

Similar problem: functions-concepts-testing-91004.html?hilit=hundreds%20function#p652379

Hope it helps.
Show more


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
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The function f is defined for each positive three-digit integer n by 04 Nov 2010, 18:24
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Bunuel
Orange08
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)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80

Let the 3-digit positive integer \({m}\) be \({abc}\) and positive integer \(v\) be \(rst\).

Then \(f(m)=2^a*3^b*5^c\) and \(f(v)=2^r*3^s*5^t\).

Given: \(f(m)=9*f(v)\) --> \(2^a*3^b*5^c=9*2^r*3^s*5^t\) --> \(2^{a-r}*3^{b-s}*5^{c-t}=3^2\) (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 \(c=t\)) --> \(m-v=20\), because if for example \(m\) and \(v\) are 143 and 123 respectively (hundreds and units digit are equal and tens digit of \(m\) is 2 more than tens digit of \(v\)) then \(143-123=20\).

Answer: D.

Similar problem: functions-concepts-testing-91004.html?hilit=hundreds%20function#p652379

Hope it helps.


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
Show more

\(2^a*3^b*5^c=2^r*3^{s+2}*5^t\): exponents of 2, 3 and 5 must be equal --> \(a=r\), \(b=s+2\) and \(c=t\).
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The function f is defined for each positive three-digit integer n by 23 Jan 2012, 12:40
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Bunuel
Orange08
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)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80

Let the 3-digit positive integer \({m}\) be \({abc}\) and positive integer \(v\) be \(rst\). Question: \((100a+10b+c)-(100r+10s+t)=?\).

Then \(f(m)=2^a*3^b*5^c\) and \(f(v)=2^r*3^s*5^t\).

Given: \(f(m)=9*f(v)\) --> \(2^a*3^b*5^c=9*2^r*3^s*5^t\) --> \(2^{a-r}*3^{b-s}*5^{c-t}=3^2\) (or \(2^a*3^b*5^c=2^r*3^{s+2}*5^t\)).

So, \(a-r=0\), \(b-s=2\) and \(c-t=0\) (or \(a=r\), \(b=s+2\) and \(c=t\)) --> \(m-v=(100a+10b+c)-(100r+10s+t)=10b-10(b-2)=20\). For example if \(m\) and \(v\) are 143 and 123 respectively (hundreds and units digit are equal and tens digit of \(m\) is 2 more than tens digit of \(v\)) then \(143-123=20\).

Answer: D.

Similar problem: functions-concepts-testing-91004.html?hilit=hundreds%20function#p652379

Hope it helps.
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Fantastic solution.
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The function f is defined for each positive three-digit integer n by 07 Mar 2015, 10:39
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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
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The function f is defined for each positive three-digit integer n by 29 Feb 2016, 13:44
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sagnik242
The function f is defined for each positive three digit integer by f(n) = 2^x 3^y 5^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) = 9f(v), then m - v = ?

A) 8
B) 9
C) 18
D) 20
E) 80

The explanation in the OG for Quant Review 2016 confused me, is it possible to pick values and solve this problem?
Show more

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.
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The function f is defined for each positive three-digit integer n by 29 Feb 2016, 21:33
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sagnik242
The function f is defined for each positive three digit integer by f(n) = 2^x 3^y 5^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) = 9f(v), then m - v = ?

A) 8
B) 9
C) 18
D) 20
E) 80

The explanation in the OG for Quant Review 2016 confused me, is it possible to pick values and solve this problem?
Show more

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
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The function f is defined for each positive three-digit integer n by 09 Aug 2017, 16:56
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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
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The function f is defined for each positive three-digit integer n by 07 Dec 2017, 14:17
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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:
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D


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The function f is defined for each positive three-digit integer n by 10 Apr 2020, 07:16
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Orange08
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 \(m-v=\) ?

(A) 8
(B) 9
(C) 18
(D) 20
(E) 80
Show more

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
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The function f is defined for each positive three-digit integer n by 09 Nov 2020, 09:46
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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.
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The function f is defined for each positive three-digit integer n by Updated on: 06 Dec 2020, 10:54
Originally posted by MHIKER on 10 Nov 2020, 00:12.
Last edited by MHIKER on 06 Dec 2020, 10:54, edited 3 times in total.
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Orange08
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 \(m-v=\) ?

(A) 8
(B) 9
(C) 18
(D) 20
(E) 80
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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
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The function f is defined for each positive three-digit integer n by 17 Mar 2021, 22:01
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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!
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The function f is defined for each positive three-digit integer n by 18 Mar 2021, 16:59
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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)]

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The function f is defined for each positive three-digit integer n by 27 Mar 2021, 05:26
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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
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The function f is defined for each positive three-digit integer n by 06 Dec 2022, 20:45
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Bunuel
Orange08
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)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80

Let the 3-digit positive integer \({m}\) be \({abc}\) and positive integer \(v\) be \(rst\). Question: \((100a+10b+c)-(100r+10s+t)=?\).

Then \(f(m)=2^a*3^b*5^c\) and \(f(v)=2^r*3^s*5^t\).

Given: \(f(m)=9*f(v)\) --> \(2^a*3^b*5^c=9*2^r*3^s*5^t\) --> \(2^{a-r}*3^{b-s}*5^{c-t}=3^2\) (or \(2^a*3^b*5^c=2^r*3^{s+2}*5^t\)).

So, \(a-r=0\), \(b-s=2\) and \(c-t=0\) (or \(a=r\), \(b=s+2\) and \(c=t\)) --> \(m-v=(100a+10b+c)-(100r+10s+t)=10b-10(b-2)=20\). For example if \(m\) and \(v\) are 143 and 123 respectively (hundreds and units digit are equal and tens digit of \(m\) is 2 more than tens digit of \(v\)) then \(143-123=20\).

Answer: D.

Similar problems:
https://gmatclub.com/forum/k-and-l-are- ... 91004.html
https://gmatclub.com/forum/the-three-di ... 63241.html
https://gmatclub.com/forum/for-any-four ... 26522.html
https://gmatclub.com/forum/k-and-l-are- ... 26646.html

Hope it helps.
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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.
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The function f is defined for each positive three-digit integer n by 07 Dec 2022, 01:28
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Bunuel
Orange08
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)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80

Let the 3-digit positive integer \({m}\) be \({abc}\) and positive integer \(v\) be \(rst\). Question: \((100a+10b+c)-(100r+10s+t)=?\).

Then \(f(m)=2^a*3^b*5^c\) and \(f(v)=2^r*3^s*5^t\).

Given: \(f(m)=9*f(v)\) --> \(2^a*3^b*5^c=9*2^r*3^s*5^t\) --> \(2^{a-r}*3^{b-s}*5^{c-t}=3^2\) (or \(2^a*3^b*5^c=2^r*3^{s+2}*5^t\)).

So, \(a-r=0\), \(b-s=2\) and \(c-t=0\) (or \(a=r\), \(b=s+2\) and \(c=t\)) --> \(m-v=(100a+10b+c)-(100r+10s+t)=10b-10(b-2)=20\). For example if \(m\) and \(v\) are 143 and 123 respectively (hundreds and units digit are equal and tens digit of \(m\) is 2 more than tens digit of \(v\)) then \(143-123=20\).

Answer: D.

Similar problems:
https://gmatclub.com/forum/k-and-l-are- ... 91004.html
https://gmatclub.com/forum/the-three-di ... 63241.html
https://gmatclub.com/forum/for-any-four ... 26522.html
https://gmatclub.com/forum/k-and-l-are- ... 26646.html

Hope it helps.

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.
Show more

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