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For any four digit number, abcd, *abcd*=

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Director
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For any four digit number, abcd, *abcd*= [#permalink] New post 22 Sep 2007, 17:37
For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What is the value of (n – m) if m and n are four-digit numbers for which *m* = (3^r)(5^s)(7^t)(11^u) and *n* = (25)(*m*)?

a) 2000
b) 200
c) 25
d) 20
e) 2

I know the answer already! Please explain with your approach, thanks.
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 [#permalink] New post 22 Sep 2007, 17:46
I have no idea how to approach this problem. where did u encounter this question.
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Subhen

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 [#permalink] New post 22 Sep 2007, 17:48
subhen wrote:
I have no idea how to approach this problem. where did u encounter this question.


one of the MGMAT practice bank questions. Some of them are pretty damn difficult. This is on the 700-800 level. =\
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 [#permalink] New post 22 Sep 2007, 19:36
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This one is really not as difficult as it seems. here is how to approach it :

From the given equation for abcd, we can clearly see that the digits of the number, abcd are the powers to which 3,5,7,11 have been raised to.

so from *m*=(3^r)(5^s)(7^t)(11^u), we know that the four digit number "m" is "rstu" and its value is 1000r+100s+10t+u

*n*=(25)(*m*) = (25)(3^r)(5^s)(7^t)(11^u)
= (3^r)(5^(s+2))(7^t)(11^u)

Thus the four digit number "n" is "r(s+2)tu" and its value is 1000r+100(s+2)+10t+u

Finally n-m = 100(s+2) - 100s = 200
  [#permalink] 22 Sep 2007, 19:36
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