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The function f is defined for each positive three-digit [#permalink]

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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 \(m-v=\) ?

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 \(f(v)=2^r*3^s*5^t\).

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 \(143-123=20\).

Re: The function f is defined for each positive three-digit [#permalink]

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04 Nov 2010, 12:40

Bunuel wrote:

Orange08 wrote:

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 \(143-123=20\).

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 \(143-123=20\).

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

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
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Re: The function f is defined for each positive three-digit [#permalink]

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23 Jan 2012, 12:40

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Bunuel wrote:

Orange08 wrote:

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 \(f(v)=2^r*3^s*5^t\).

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 \(143-123=20\).

Re: The function f is defined for each positive three-digit [#permalink]

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28 Mar 2014, 07:03

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Re: The function f is defined for each positive three-digit [#permalink]

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07 Mar 2015, 10:39

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

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

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.

As for how to learn about functions, my this week's and next week's blog posts will deal with functions so check them out on the Veritas blog.
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Re: The function f is defined for each positive three-digit [#permalink]

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18 Jul 2015, 13:19

Orange08 wrote:

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

Please change to "then". Thank you
_________________

Re: The function f is defined for each positive three-digit [#permalink]

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29 Feb 2016, 13:29

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?

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?

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.

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.

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?

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

Re: The function f is defined for each positive three-digit [#permalink]

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21 Oct 2016, 07:03

Orange08 wrote:

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

f(m) = 9f(v) 9f(v) = 2^x * 3^2y ^ 5^z so the difference here is that the power of m for 3 is 2y since xyz represent the digits of the number, we can have few options. y for v can be 1,2,3,4,5. y for m can be 2,4,6,8,0

since y stands for tens digit, and z stand for units digit, it must be true that z is the units digit for both, therefore A, B, and C are not possible, as the difference would be smth to _0, where _ stands for tens digit of the difference. for tens digit of the difference, we can have: 1, 2, 3, 4, 5. the difference thus might be: 10, 20, 30, 40, 50. as we can see, only D fits!

Re: The function f is defined for each positive three-digit [#permalink]

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12 May 2017, 21:52

Orange08 wrote:

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

To solve this question simply reverse engineer- pick a random value for f(v) such as 123

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