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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Brian@VeritasPrep
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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!
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
Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
GMAT Instructor
Chief Academic Officer
Veritas Prep
Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
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anshumishra
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5^n > 4*10^7GHong14 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 > 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
(Every mistake is a lesson learned )
Anshu
(Every mistake is a lesson learned )
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Night reader
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4*(10^7)=40,000,000GHong14 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
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
