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Re: M02-16 [#permalink]
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Bunuel wrote:
Which of the following is the smallest number?

A. \(\frac{1}{900 * 5^4}\)
B. \(\frac{1}{80 * 5^5}\)
C. \(\frac{1}{5000 * 5^3}\)
D. \(\frac{1}{40 * 5^6}\)
E. \(\frac{1}{30000 * 5^2}\)


Bunuel's method is probably the quickest way to solve, but to avoid making mistakes, I broke the denominators down into prime factors - it just requires a bit of focus to make sure to count all the 0's properly, as these represent factors of 2 and 5.

The highest denominator tells us the smallest number. So let's figure out the denominators.

A. \(\frac{1}{900 * 5^4} --> 2^2*3^2*5^6\)
B. \(\frac{1}{80 * 5^5} --> 2^4*5^6\)
C. \(\frac{1}{5000 * 5^3} --> 2^3*5^7\)
D. \(\frac{1}{40 * 5^6} --> 2^3*5^7\)
E. \(\frac{1}{30000 * 5^2} --> 2^4*3*5^6\)

Why did I choose E?

Well, 5^7 is obviously much greater than 5^6. But, instead of having the greatest exponent of 5, E has an extra factor of 2*3 (6).
Thus, \(5^6 * 6 > 5^7\)
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Re: M02-16 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M02-16 [#permalink]
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