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# What's the easiest way to solve this?

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Intern
Joined: 15 Oct 2009
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What's the easiest way to solve this? [#permalink]

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27 Mar 2010, 14:59
00:00

Difficulty:

(N/A)

Question Stats:

100% (00:08) correct 0% (00:00) wrong based on 6 sessions

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If 5^21 * 4^11 = 2 * 10^n what is the value of n?

a) 11
b) 21
c) 22
d) 23
e) 32

Kudos [?]: 154 [0], given: 11

Senior Manager
Joined: 21 Jul 2009
Posts: 364

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Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.
Re: What's the easiest way to solve this? [#permalink]

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27 Mar 2010, 15:07
Ans b.
rewrite the question statement as - 5^21 * 2^21 * 2
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Kudos [?]: 200 [0], given: 22

Senior Manager
Joined: 01 Feb 2010
Posts: 251

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Re: What's the easiest way to solve this? [#permalink]

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27 Mar 2010, 23:18
changhiskhan wrote:
If 5^21 * 4^11 = 2 * 10^n what is the value of n?
a) 11
b) 21
c) 22
d) 23
e) 32

5^21 * 2^(2*11) = 2*10^n
5^21 * 2^22 = 2*10^n
10^21 * 2 = 2*10^n (Multiplying 2^21 with 5^21)
so n = 21 hence B

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Manager
Joined: 26 Feb 2010
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Location: Argentina
Re: What's the easiest way to solve this? [#permalink]

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29 Mar 2010, 12:04
changhiskhan wrote:
If 5^21 * 4^11 = 2 * 10^n what is the value of n?

a) 11
b) 21
c) 22
d) 23
e) 32

Rewriting some of the terms:
$$4^{11} = (2*2)^{11} = 2^{22} 2*10^n = 2*(2*5)^n = 2*2^n*5^n= 2^{(n+1)}$$if equal the five to five and two with two
$$5^{21} = 5^n$$and $$2^{22} = 2^{(n+1)}$$

So n must be 21

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Intern
Joined: 23 Feb 2010
Posts: 14

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Re: What's the easiest way to solve this? [#permalink]

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29 Mar 2010, 22:56
$$5^{21}*4^{11} = 2*10^n$$

Taking a "5" and a "4" out of their exponent sets lets me get to "2 x (2)(5)" which is starting to look like what I need:
$$5*5^{20}*4*4^{10} = 2*10^n$$
$$2*2*5*5^{20}*4^{10} = 2*10^n$$
$$2*10*5^{20}*4^{10} = 2*10^n$$

With what's left, there are two 2s in every 4. This gives me twenty 2s and twenty 5s - letting me make twenty 10s through multiplication:

$$2*10*10^{20} = 2*10^n$$

Reduce and solve for n:
$$2*10^{21} = 2*10^n$$
$$n = 21$$ (b)

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CEO
Joined: 17 Nov 2007
Posts: 3583

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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: What's the easiest way to solve this? [#permalink]

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29 Mar 2010, 23:00
Another way:

The number of "5"s on the left side has to equal that on the right side: $$5^{21} = 5^n$$ --> n = 21
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Kudos [?]: 4672 [0], given: 360

Re: What's the easiest way to solve this?   [#permalink] 29 Mar 2010, 23:00
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