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by aleph777 » Wed Jan 12, 2011 2:35 pm
GHong14,

Not sure if there's an 'easy' way to do this one. It's a prime factorization problem to me.

Basically, it's asking, how many 5s are in 40,000,000. And since 40,000,000 is such a nice round number, you can start breaking it down into factors of 5.

40,000,000/5 = 8,000,000
8,000,000/5 = 1,600,000
1,600,000/5 = 320,000
320,000/5 = 64,000
64,000/5 = 12,800
12,800/5 = 2,560
2,560/5 = 512

And now it gets a little tricky since we can't evenly divide any longer.

But since we know 5^4 = 625 and 625 is slightly larger than 512 in the grand scheme, you can stop factorizing and just count up your fives.

Answer: E
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by Brian@VeritasPrep » Wed Jan 12, 2011 3:10 pm
Great explanation, aleph777 - I was thinking pretty similarly. Since 5^n is a prime base to an exponent, I want to try to make 40,000,000 similar:

40,000,000 = 4 * 10^7

4 * 10^7 = 4 * 2^7 * 5^7 (now we have a common 5^exponent in there)

= 2^9 * 5^7

Now that at least the 5^7 term is in the same form as the 5^n term, it's a little easier to compare:

5^n > 5^7 * 2^9

So...we can divide both sides by 5^7:

5^(n-7) > 2^9


That's the algebraic way, or you can just think it through conceptually; we're comparing:

a) 5^7 > 5^7 * 2^9 ? (clearly, no - you can divide both sides by 5^7 and 1 is not greater than 2^9)

b) 5^8 > 5^7 * 2^9? (no - 5 is not greater than 2^9)

c) 5^9 > 5^7 * 2^9?
divide both sides by 5^7: 5^2 > 2^9 (no)

d) 5^10 > 5^7 * 2^9?
divide both sides by 5^7: 5^3 > 2^9? (no - you don't even really need to fully calculate 2^9, either - just start multiplying 2s: 2, 4, 8, 16, 32, 64, 128 - by 2^7 we're already done)

e) 5^11 > 5^7 * 2^9? NOTE: e is all we have left, so it's correct!

divide both sides:

5^4 > 2^9?
when we left off at 2^7 the two were just about equal (125 vs. 128); now we're multiplying the left side by 5 and the right side by 4 (to get all the way to 2^9). Therefore, the left will be bigger, and E is correct.



Now, the explanation may be a bit labor-intensive, but the idea of finding common bases for the 5 should get you pretty quickly on your way and you may be able to do a lot of this in your head or with some quick scratchwork from there. When in doubt with exponents, prime-factor big numbers to find common bases!
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep

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by anshumishra » Wed Jan 12, 2011 3:12 pm
GHong14 wrote:If n is an integer, what is the least possible value of n such that 40,000,000 < 5^n ?

a 7
b 8
c 9
d 10
e 11
5^n > 4*10^7
=> 5^n > 2^2*2^7*5^7
=>5^(n-7) > 2^9
=>5^(n-7) > 512
So, the minimum value of (n-7) = 4
=> n = 11 E
Thanks
Anshu

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by Night reader » Wed Jan 12, 2011 3:18 pm
GHong14 wrote:If n is an integer, what is the least possible value of n such that 40,000,000 < 5^n ?

a 7
b 8
c 9
d 10
e 11
4*(10^7)=40,000,000
OR
4* ((2*5)^7) for easy calculation => 2^7= 2^5 * 2^2 OR 32*4=148 :)
4*148*(5^7) [not ready yet :) ] 4*148 => 400+160+32=592 [which is less than 625 OR 25^2, remember useful squares :) ]
so 592 < (5^2)^2 OR 592 < 5^4
what we have? 5^7, 4*148=592 => (5^7)*(5^4) should be greater than 4*148*(5^7), isn't it?
so answer is n=(7+4) for (5^7)*(5^4)=5^(7+4), n=11
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