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5^21 * 4*11 = 2 * 10^n ..... what is n?

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Junior | Next Rank: 30 Posts
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by samus33 » Mon Dec 07, 2009 7:37 am
Hey Everyone,

What is the most efficient way to approach this problem? Because it was one of the first questions I was handed on GMAT CAT 1, I spent time to actually write out the problem. It was terribly unpleasant, and inefficient.

Can someone propose a way to attack it more agressively?

Cheers
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by bkholy » Mon Dec 07, 2009 8:02 am
Are you sure you setup the problem right? I think it should be, (5^21)*(4^11)=2(10^n)
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by JasLamba » Mon Dec 07, 2009 9:06 am
Hey,

Whenever you see exponents it is good to factorize everything and put it in prime format.

so 5^21 * (2*2)^11 = 2*(5*2)^N
from the exponent rules of addition - you can now simplify to:
5^21 * 2^11*2^11= 2^1*2^N*5^N
automatically from here you can see that N= 21 as both sides have to be equal and we can know this from the 5. If you want to really clean things up you can simplify the 2's - (divide 2^22/2^1) and get a final equation of:
5^21 *2^21 = 5^N * 2^N

hope it helped.
Jas
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