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The function f is defined for each positive threedigit [#permalink]
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The function f is defined for each positive threedigit 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 threedigit positive integers such that \(f(m)=9*f(v)\) , then \(mv=\) ? (A) 8 (B) 9 (C) 18 (D) 20 (E) 80
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Re: The function f is defined for each positive threedigit [#permalink]
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f(m)=9f(v) can be written as f(m) = \(3^2\)f(v). So tenth place of m is 2 number greater than tenth place of v. This means m is 20 greater than v. Answer D
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Orange08 wrote: The function f is defined for each positive threedigit 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 threedigit positive integers such that f(m)=9f(v), them mv=? (A) 8 (B) 9 (C) 18 (D) 20 (E) 80 Let the 3digit 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: \(f(m)=9*f(v)\) > \(2^a*3^b*5^c=9*2^r*3^s*5^t\) > \(2^{ar}*3^{bs}*5^{ct}=3^2\) (or \(2^a*3^b*5^c=2^r*3^{s+2}*5^t\)). So, \(ar=0\), \(bs=2\) and \(ct=0\) (or \(a=r\), \(b=s+2\) and \(c=t\)) > \(mv=(100a+10b+c)(100r+10s+t)=10b10(b2)=20\). 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 \(143123=20\). Answer: D. Similar problems: kandlareeachfourdigitpositiveintegerswiththousands91004.htmlthethreedigitpositiveintegerxhasthehundreds163241.htmlforanyfourdigitnumberabcdabcd3a5b7c11d126522.htmlkandlareeachfourdigitpositiveintegerswiththousands126646.htmlHope it helps.
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Re: The function f is defined for each positive threedigit [#permalink]
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04 Nov 2010, 11:40
Bunuel wrote: Orange08 wrote: The function f is defined for each positive threedigit 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 threedigit positive integers such that f(m)=9f(v), them mv=? (A) 8 (B) 9 (C) 18 (D) 20 (E) 80 Let the 3digit 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: \(f(m)=9*f(v)\) > \(2^a*3^b*5^c=9*2^r*3^s*5^t\) > \(2^{ar}*3^{bs}*5^{ct}=3^2\) (or \(2^a*3^b*5^c=2^r*3^{s+2}*5^t\)) > \(ar=0\), \(bs=2\) and \(ct=0\) (or \(a=r\), \(b=s+2\) and \(c=t\)) > \(mv=20\), 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 \(143123=20\). Answer: D. Similar problem: functionsconceptstesting91004.html?hilit=hundreds%20function#p652379Hope 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 \(2^a*3^b*5^c=9*2^r*3^s*5^t\) > \(2^{ar}*3^{bs}*5^{ct}=3^2\) (or \(2^a*3^b*5^c=2^r*3^{s+2}*5^t\)) > \(ar=0\), \(bs=2\) and \(ct=0\) (or \(a=r\), \(b=s+2\) and \(c=t\)) how do we reach ar=0 ? because both a and r are on different side with base 2
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Re: The function f is defined for each positive threedigit [#permalink]
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04 Nov 2010, 17:24
hirendhanak wrote: Bunuel wrote: Orange08 wrote: The function f is defined for each positive threedigit 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 threedigit positive integers such that f(m)=9f(v), them mv=? (A) 8 (B) 9 (C) 18 (D) 20 (E) 80 Let the 3digit 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: \(f(m)=9*f(v)\) > \(2^a*3^b*5^c=9*2^r*3^s*5^t\) > \(2^{ar}*3^{bs}*5^{ct}=3^2\) (or \(2^a*3^b*5^c=2^r*3^{s+2}*5^t\)) > \(ar=0\), \(bs=2\) and \(ct=0\) (or \(a=r\), \(b=s+2\) and \(c=t\)) > \(mv=20\), 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 \(143123=20\). Answer: D. Similar problem: functionsconceptstesting91004.html?hilit=hundreds%20function#p652379Hope 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 \(2^a*3^b*5^c=9*2^r*3^s*5^t\) > \(2^{ar}*3^{bs}*5^{ct}=3^2\) (or \(2^a*3^b*5^c=2^r*3^{s+2}*5^t\)) > \(ar=0\), \(bs=2\) and \(ct=0\) (or \(a=r\), \(b=s+2\) and \(c=t\)) how do we reach ar=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 \(c=t\).
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Re: The function f is defined for each positive threedigit [#permalink]
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Orange08 wrote: The function f is defined for each positive threedigit 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 threedigit positive integers such that f(m)=9f(v), them mv=? (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 threedigit integer n by \(f(n) = 2^x3^y5^z\) , where x, y and z are the hundreds, tens, and units digits of n, respectivelyI 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 threedigit [#permalink]
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Bunuel wrote: Orange08 wrote: The function f is defined for each positive threedigit 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 threedigit positive integers such that f(m)=9f(v), them mv=? (A) 8 (B) 9 (C) 18 (D) 20 (E) 80 Let the 3digit 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: \(f(m)=9*f(v)\) > \(2^a*3^b*5^c=9*2^r*3^s*5^t\) > \(2^{ar}*3^{bs}*5^{ct}=3^2\) (or \(2^a*3^b*5^c=2^r*3^{s+2}*5^t\)). So, \(ar=0\), \(bs=2\) and \(ct=0\) (or \(a=r\), \(b=s+2\) and \(c=t\)) > \(mv=(100a+10b+c)(100r+10s+t)=10b10(b2)=20\). 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 \(143123=20\). Answer: D. Similar problem: functionsconceptstesting91004.html?hilit=hundreds%20function#p652379Hope it helps. Fantastic solution.
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Re: The function f is defined for each positive threedigit [#permalink]
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20 Apr 2014, 07:49
m=xyz v=pqr mv = 100(xp) + 10(yq) + (zr)............................(1) f(m)= 2^x*3^y*5^z f(v)= 2^p*3^q*5^r Given that, f(m) = 9 f(v) => 2^x*3^y*5^z = 9* 2^p*3^q*5^r => 2^x*3^y*5^z = 3^2*2^p*3^q*5^r => 2^x*3^y*5^z = 2^p*3^(q+2)*5^r => 2^[xp]*3^[yq2]*5^[zr] = 1 => 2^[xp]*3^[yq2]*5^[zr] = 2^0*3^0*5^0 (xp)=0 (zr)=0 (yq)=2 Substituting the above values in equation (1). mv = 100(xp) + 10(yq) + (zr) => 0 + 10*2 + 0 =>20
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Re: The function f is defined for each positive threedigit [#permalink]
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07 Mar 2015, 09:39
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



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Re: The function f is defined for each positive threedigit [#permalink]
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erikvm wrote: 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. 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 threedigit [#permalink]
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18 Jul 2015, 20:21
f(m) = 2^x 3^y 5^z 9f(v) = 9(2^x 3^y 5^z) = 2^x 3^(y+2) 5^z so, f(m) = 9f(v) since the tens place of f(m) is 2 more than the tens place of f(v), mv = 20. Ans (D).
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Re: The function f is defined for each positive threedigit [#permalink]
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sagnik242 wrote: 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.If you want the algebraic solution, refer to Bunuel 's post above thefunctionfisdefinedforeachpositivethreedigit100847.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 mv=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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Re: The function f is defined for each positive threedigit [#permalink]
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29 Feb 2016, 20:33
sagnik242 wrote: 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.. mv= 100x+10y+20+z(100x+10y+z )=20
ans 20 D
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Re: The function f is defined for each positive threedigit [#permalink]
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21 Oct 2016, 06:03
Orange08 wrote: The function f is defined for each positive threedigit 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 threedigit positive integers such that f(m)=9f(v), them mv=? (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!



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Re: The function f is defined for each positive threedigit [#permalink]
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06 Feb 2017, 21:01
f(n) = 2^x * 3^y * 5^z
suppose f(m) = 2^A * 3^B * 5^C and f(v) = 2^p * 3^q * 5^r
as per question stem => f(m) = 9 f(v) => 2^A *3^B * 5^C = 3^2 (2^p * 3^q * 5^r) => 2^A *3^B * 5^C = (2^p * 3^(q+2) * 5^r) so A = p ; B = q+2 ; C = r
now => mv => (100A + 10B + C)  (100p + 10q + r) => (100p + 10(q+2) + r)  100p  10q  r => 20 Ans



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Re: The function f is defined for each positive threedigit [#permalink]
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23 Mar 2017, 04:09
from the infor we see that f(m)=9f(v) if we adjust exponenst we will have 2^x * 3^y * 5^z=2^x * 3^(y+2) * 5 ^z
x y z minus x (y+2) z y(y+2) yy=2 as y is the tenths, then the difference is 20



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Re: The function f is defined for each positive threedigit [#permalink]
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12 May 2017, 20:52
Orange08 wrote: The function f is defined for each positive threedigit 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 threedigit positive integers such that \(f(m)=9*f(v)\) , then \(mv=\) ?
(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 F(v) = 2^x3^y5^z f(123)= 2^1 3^2 5^3 f(m) = 2^1 3^2 5^3 * 9 f(m) = 2^1 3^2 5^3 * 3^2 f(m) = 2^1 3^4 5^3 143123 = 20



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The function f is defined for each positive threedigit [#permalink]
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09 Aug 2017, 15:56
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
120100=20
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Re: The function f is defined for each positive threedigit [#permalink]
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01 Sep 2017, 08:30
Hi Bunuel,
My question here is how you can write 2^a x 3^b x 5^c = 100a + 10b + c. Is this a property, if yes then why 120 three digit positive integer is not as per this property.



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Re: The function f is defined for each positive threedigit [#permalink]
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15 Oct 2017, 07:18
Orange08 wrote: The function f is defined for each positive threedigit 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 threedigit positive integers such that \(f(m)=9*f(v)\) , then \(mv=\) ?
(A) 8 (B) 9 (C) 18 (D) 20 (E) 80 2^x3^y5^z is the function: If you caught the above highlighted clue from the question stem you are through. Since the differential is of 3^2 means that at the tens position, 10*2 is the additional figure to the overall number, m. Thus, MV=10(x+2)10x = 20




Re: The function f is defined for each positive threedigit
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15 Oct 2017, 07:18



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