IMO D

A \(\frac{3^2}{7^5*5^3}\) = \(\frac{9}{7^5*5^3}\)

B \(\frac{2^2}{7^5 * 5^3}\) = \(\frac{4}{7^5 * 5^3}\)

C \(\frac{3^3}{7^6 * 5^3}\) = \(\frac{27}{7^6 * 5^3}\)

D \(\frac{21}{7^5 * 5^3}\) = \(\frac{21}{7^5 * 5^3}\)

E \(\frac{3^2 * 2}{7^6 * 5^3}\)[/quote] = \(\frac{18}{7^6 * 5^3}\)[/quote]

Now if you can see we have two sets of fractions in which

\(\frac{21}{7^5 * 5^3}\) & \(\frac{27}{7^6 * 5^3}\)

Out of which

\(\frac{21}{7^5 * 5^3}\)
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Use splits.

If all else is equal, a smaller denominator creates a fraction with greater value (more to the right on the number line).

\(\frac{1}{3}<\frac{1}{2}\)

All denominators in the options contain \(5^3\)

But C and E contain \(7^6\), while the others contain \(7^5\)
-- \(7^6\) is much greater than \(7^5\).
-- We need a smaller denominator

Eliminate C and E

Now we have the same denominator, and its value does matter. The denominator could equal 100, or 1,000,000.

In fact, we can write the denominator as "100" for easy comparison.

We need the greatest numerator.

A greater numerator over the same denominator equals a fraction with greater value.
(\(\frac{9}{10}>\frac{3}{10}\))

Numerators:

A) \(3^2=9\)
\(\frac{9}{100}=0.09\)

B) \(2^2=4\)
\(\frac{4}{100}=0.04\)

D) \(3^3=27\)
\(\frac{27}{100}=0.27\)



Answer
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Bunuel, I used cross multiplication to compare A and D. Is there a better method?

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