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m^a * n^b = 200, all +ve integers. a+b = ? A. a > b B. a

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Manager
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m^a * n^b = 200, all +ve integers. a+b = ? A. a > b B. a [#permalink]  15 May 2008, 17:27
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m^a * n^b = 200, all +ve integers. a+b = ?

A. a > b
B. a * b =6
Senior Manager
Joined: 20 Feb 2008
Posts: 296
Location: Bangalore, India
Schools: R1:Cornell, Yale, NYU. R2: Haas, MIT, Ross
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Kudos [?]: 38 [0], given: 0

Re: DS: A Simple Question?? [#permalink]  15 May 2008, 17:31

m^a *n^b =200
Breaking 200 to its prime roots we get
m^a *n^b =2^3 *5^2

Both statments satisfy the condition
Director
Joined: 23 Sep 2007
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Re: DS: A Simple Question?? [#permalink]  15 May 2008, 17:42
B

m^a *n^b =200
Breaking 200 to its prime roots we get
m^a *n^b =2^3 *5^2

But you can also have a situation where m^a *n^b =2^2 *50^1
so statement is not suff

statement 2
a*b = 6
a = 2, b = 3 or a = 3, b = 2
OR
a = 1, b = 6 or a = 6, b = 1 but this is not possible if you break down 200

therefore a*b = 6 is suff
Manager
Joined: 04 Sep 2007
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Re: DS: A Simple Question?? [#permalink]  15 May 2008, 17:47
gmatnub wrote:
B

m^a *n^b =200
Breaking 200 to its prime roots we get
m^a *n^b =2^3 *5^2

But you can also have a situation where m^a *n^b =2^2 *50^1
so statement is not suff

statement 2
a*b = 6
a = 2, b = 3 or a = 3, b = 2
OR
a = 1, b = 6 or a = 6, b = 1 but this is not possible if you break down 200

therefore a*b = 6 is suff

Thanks, mate. That's a neat way to solve.
Senior Manager
Joined: 20 Feb 2008
Posts: 296
Location: Bangalore, India
Schools: R1:Cornell, Yale, NYU. R2: Haas, MIT, Ross
Followers: 4

Kudos [?]: 38 [0], given: 0

Re: DS: A Simple Question?? [#permalink]  15 May 2008, 18:56
Good job, needed to watch out for those kind of errors!
Re: DS: A Simple Question??   [#permalink] 15 May 2008, 18:56
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