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Fundemental Exponent Question

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Fundemental Exponent Question [#permalink]

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New post 21 Jul 2009, 13:33
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A
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C
D
E

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\(2^5+2^5+3^5+3^5+3^5\) is equal to?

A. \(5^6\)
B. \(13^{15}\)
C. \(2^6+3^6\)
D. \(2^7+3^8\)
E. \(4^5+9^{15}\)

I know the answer is C, but I can only find this out by solving both. There must be a faster way?

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Senior Manager
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Schools: Duke,Oxford,IMD,INSEAD
Re: Fundemental Exponent Question [#permalink]

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New post 21 Jul 2009, 14:19
(2^5)+(2^5)+(3^5)+(3^5)+(3^5)

Take common factors out ==> (2^5)(1+1) + (3^5)(1+1+1) = (2^5)*2 + (3^5)*3 = (2^6)+(3^6) therefore answer is C

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Re: Fundemental Exponent Question [#permalink]

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New post 21 Jul 2009, 16:09
Think of 2^5 and 3^5 as x and y.

Thus you have x+x+y+y+y, which means 2x + 3y.

Now returning to 2^5 and 3^5, you have
2x + 3y
2*2^5 + 3*3^5

applying the properties, maintain the base and sum the exponents.
2^(1+5) + 3^(1+5)
2^6 + 3^6

PS.: It worked? consider a kudo ;)

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Re: Fundemental Exponent Question [#permalink]

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New post 31 Jul 2009, 11:29
(2^5)+(2^5) = 2(2^5) = 2(2*2*2*2*2) = 2^6

Same to the 3's:

(3^5)+(3^5)+(3^5) = 3(3^5) = 3(3*3*3*3*3) = 3^6

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Re: Fundemental Exponent Question [#permalink]

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New post 01 Aug 2009, 10:47
2^5+2^5+3^5+3^5+3^5
=2* (2^5) + 3*(3^5)
=[2^(1+5)] +[3^(1+5)]
=(2^6) + (3^6)

Kudos [?]: 19 [0], given: 1

Re: Fundemental Exponent Question   [#permalink] 01 Aug 2009, 10:47
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