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Arithmetic Exponents

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by sk02 » Tue Nov 10, 2009 10:42 pm
Not sure if this is the easiest answer, but if you group the 2s and 3s, it should help to identify the ans:

(2^5+2^5)+(3^5+3^5+3^5)
= 2(2^5) + 3(3^5)
= 2^6 + 3^6

so ans C.[/list]
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by heshamelaziry » Tue Nov 10, 2009 11:01 pm
C it is
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by antondesh » Wed Nov 11, 2009 6:39 am
Algebraically, it solves like this:

2^5+2^5+3^5+3^5+3^5 =

2^5*(1+1)+3^5*(1+1+1)=

2^5*2^1+3^5*3^1=

2^6+3^6

Choice C.
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