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m - v = ?

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by mehravikas » Mon Dec 14, 2009 12:00 pm
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), then m-v = ?

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

OA : D
Source: GMAT Sets

I am getting C
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by mbaonmind » Mon Dec 14, 2009 7:37 pm
If m and v are three-digit positive integers
Let m = xyz
v = abc
since f(m) = 9 f(v)
2^x 3^y 5^z = 9(2^a 3^b 5^c)
= 2^(x-a) 3^(y-b) 5^(z-c) = 9

since 9 is a multiple of 3 we can deduce that x-a = 0
y-b = 2
z-c = 0

hence m-v = 20 :)
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by mehravikas » Tue Dec 15, 2009 12:17 pm
Can we plug in numbers for this problem?
mbaonmind wrote:If m and v are three-digit positive integers
Let m = xyz
v = abc
since f(m) = 9 f(v)
2^x 3^y 5^z = 9(2^a 3^b 5^c)
= 2^(x-a) 3^(y-b) 5^(z-c) = 9

since 9 is a multiple of 3 we can deduce that x-a = 0
y-b = 2
z-c = 0

hence m-v = 20 :)
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by mbaonmind » Tue Dec 15, 2009 4:49 pm
sure you take up any number that differ by 20
e.g 111 and 131 or
125 and 145

using above formulae f(m) = 9f(v) always where m < v
[/quote]
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by sanju09 » Wed Dec 16, 2009 1:49 am
Accordingly, let f(m) = 2^a 3^b 5^c and f(v) = 2^d 3^e 5^f such that 2^a 3^b 5^c = 9 × (2^d 3^e 5^f); or 2^(a-d) 3^(b-e) 5^(c-f) = 3^2, we otherwise have

a-d = 0, b-e = 2, and c-f = 0

now say that we have two 3-digit positive integers such that their hundred's and unit's digits are same, but the ten's digit of the greater integer is 2 more than that of the smaller; just like this example for m and v:

m = 259 and v = 239 and m-v = 20. Take D
The mind is everything. What you think you become. -Lord Buddha



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by mehravikas » Fri Dec 18, 2009 1:29 pm
Why do we have to pick numbers that have a difference of 20?
sanju09 wrote:Accordingly, let f(m) = 2^a 3^b 5^c and f(v) = 2^d 3^e 5^f such that 2^a 3^b 5^c = 9 × (2^d 3^e 5^f); or 2^(a-d) 3^(b-e) 5^(c-f) = 3^2, we otherwise have

a-d = 0, b-e = 2, and c-f = 0

now say that we have two 3-digit positive integers such that their hundred's and unit's digits are same, but the ten's digit of the greater integer is 2 more than that of the smaller; just like this example for m and v:

m = 259 and v = 239 and m-v = 20. Take D
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by sanju09 » Sat Dec 19, 2009 12:31 am
mehravikas wrote:Why do we have to pick numbers that have a difference of 20?


Since the digit's at ten's place differ by two, with the other digits being the same, the two 3-digit positive integers would differ by [spoiler]2 times 10 = 20[/spoiler].
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by mehravikas » Sat Dec 19, 2009 3:00 pm
thanks for this :-)
sanju09 wrote:
mehravikas wrote:Why do we have to pick numbers that have a difference of 20?


Since the digit's at ten's place differ by two, with the other digits being the same, the two 3-digit positive integers would differ by [spoiler]2 times 10 = 20[/spoiler].
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