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good one

by vaivish » Tue Jul 29, 2008 11:23 am
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
(C)18
(D)20
(E)80
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Re: good one

by Ian Stewart » Tue Jul 29, 2008 3:06 pm
vaivish wrote: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
(C)18
(D)20
(E)80
I'm guessing you mean: f(x) = 2^x * 3^y * 5^z. Otherwise there is not a unique answer here. If f(m) = 9f(v) = 3^2 f(v), then the exponent on the 3 must be two greater in f(m) than in f(v) while the other exponents must be the same, which means the tens digit is two greater in m than in v, so the two numbers must differ by 20.

[edited- I had written m and v where I meant f(m) and f(v).]
Last edited by Ian Stewart on Tue Jul 29, 2008 5:18 pm, edited 1 time in total.
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by oldgeezer » Tue Jul 29, 2008 3:17 pm
I think it is A, 8. IF you meant it like written, 2x3y5z, then the lowest possible number is 111 (any zeros would make the function 0 and therefore nothing could be 9 times its value)

the function of 111 is 30

I chose this number to start because the values of the answers are not that large and making the numbers much bigger quickly eliminates all answers

Anyway, since the value of the function of the other number must be 9 times 30 or 270, I came up with the number 119, since the value of its function is 270 (2*3*45) and the difference with 111 is 8.

I don't think there are any other combinations of numbers that satisfy the 9x function value and any of the answers

OF COURSE, this took more than 2 minutes!!! and this is what kills me about this test

old geezer
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by Ian Stewart » Tue Jul 29, 2008 5:16 pm
oldgeezer wrote:I think it is A, 8. IF you meant it like written, 2x3y5z, then the lowest possible number is 111 (any zeros would make the function 0 and therefore nothing could be 9 times its value)

the function of 111 is 30

I chose this number to start because the values of the answers are not that large and making the numbers much bigger quickly eliminates all answers

Anyway, since the value of the function of the other number must be 9 times 30 or 270, I came up with the number 119, since the value of its function is 270 (2*3*45) and the difference with 111 is 8.

I don't think there are any other combinations of numbers that satisfy the 9x function value and any of the answers

OF COURSE, this took more than 2 minutes!!! and this is what kills me about this test

old geezer
That's all correct, but notice that if we accept the question as written, and v = 111, then m could be 119, 191, 911, 331, 313 or 133. If m is 191, then m-v would be 80, not 8. That's why I was quite sure that the question was not typed correctly- there would be more than one correct answer if it were. I'd also expect to see "2x3y5z" written differently- it's just equal to 30xyz (and the 30 is irrelevant to the question).

If x, y and z are actually exponents, then there is only one correct answer.
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by oldgeezer » Tue Jul 29, 2008 5:40 pm
very good! thank you. I couldn't find any other number combination that would satisify the terms of the question and the given answer choices (I missed the 191 and 80)

Then again, once I found one I didn't look much further... I am actually shocked there is another combination BY CHANCE when the problem is written incorrectly... or does this show how far the GMAT people will go to find answers you might get to the wrong way? you woul dhave to missinterpret the exponents for factors... wow do they go THAT far in searching for the "right" wrong answers?

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by umaa » Wed Jul 30, 2008 1:29 am
OA pls.
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by Ian Stewart » Wed Jul 30, 2008 3:34 am
oldgeezer wrote: I am actually shocked there is another combination BY CHANCE when the problem is written incorrectly... or does this show how far the GMAT people will go to find answers you might get to the wrong way? you woul dhave to missinterpret the exponents for factors... wow do they go THAT far in searching for the "right" wrong answers?
Well, I'm not sure where the question in the above post is from, but yes, you might be surprised by how much thought goes into writing the wrong answer choices for real GMAT questions. If you analyze the answer choices to almost any appropriate GMAT question, you can normally work out why most of them are there- that is, work out what common mistake the question designer has anticipated. Just choosing a random example from the OG (qn 126), and simplifying the language to save me some typing (:)):

Machine A does a job in 3 hours. Machine B does the same job in 4 hours. How long will it take A and B working together?

7/12
1 1/2
1 5/7
3 1/2
7

Answer A is the reciprocal of the correct answer (and if someone use the rates formula in this form: 1/a + 1/b = 1/t, it requires you to take the reciprocal at the end; someone who forgets to do this will arrive at 7/12). Answer D is the average of 3 and 4, while answer E is the sum of 3 and 4 - two simple arithmetic combinations someone might arrive at if they misunderstood the question. B doesn't seem to come from any common arithmetic error, but it is very close in value to the correct answer, so may be there to trap someone who estimates poorly.

Incidentally, the question can be done very quickly: two machines like A would do the job in 3/2 hours, but B is slower than A. It must take longer than 3/2 hours. Two machines like B would do the job in 4/2 = 2 hours, but A is faster: the correct answer must be between 3/2 and 2.
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by vishubn » Thu Oct 02, 2008 8:31 pm
Hi Ian !!

I didn uderstand this part
Otherwise there is not a unique answer here. If f(m) = 9f(v) = 3^2 f(v), then the exponent on the 3 must be two greater in f(m) than in f(v)
couuld u please expain with much easier examples

Vishu
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by vaivish » Sat Oct 04, 2008 2:41 am
hey thanks a ton...but the question posted is right.....i have checked it again...but forgot the source.....
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by Fab » Mon Oct 13, 2008 7:06 pm
Could someone please elaborate more?

THANKS
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by Ian Stewart » Tue Oct 14, 2008 1:20 am
vaivish wrote:hey thanks a ton...but the question posted is right.....i have checked it again...but forgot the source.....
Hi vaivish- it's not your fault, obviously, but whoever published the question didn't print it correctly. The x, y and z are supposed to be exponents. So the question should read:

The function f is defined for each positive three-digit integer n 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), them m-v=?

I gave a brief solution above, but in further detail:

if m = ABC, where A, B and C are digits, and v = DEF, where D, E and F are digits, then

f(m) = 2^A * 3^B * 5^C

and

f(v) = 2^D * 3^E * 5^F

Since f(m) = 9f(v), we have:

2^A * 3^B * 5^C = 9 * 2^D * 3^E * 5^F
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

and by the Fundamental Theorem of Arithmetic (unique factorization into primes), if two integers are equal, they have the same prime factorizations, so the powers on the left and on the right must be equal:

A = D
B = E+2
C = F

So the number ABC is only different from DEF in the tens digit- its tens digit is two greater, so ABC is 20 larger than DEF, and m is 20 larger than v.
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