AMC 10 Step by Step

2013 AMC 10A Problem 8

P1-10~1 minPrint

What is the value of 22014+2201222014−22012 ?\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?

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Key insight

Write 2 to the 2014 as 4 times 2 to the 2012, pull 2 to the 2012 out of top and bottom, and 5 over 3 is left.

Solution

The powers are huge, so the actual numbers cannot matter. Only how they compare matters.

1. Try small numbers first. Replace 20142014 and 20122012 by 44 and 22:
24+2224−22=16+416−4=2012=53.\frac{2^4 + 2^2}{2^4 - 2^2} = \frac{16 + 4}{16 - 4} = \frac{20}{12} = \frac53.
Notice 1616 is 44 times 44. The top is 4×(4+1)4 \times (4 + 1) and the bottom is 4×(4−1)4 \times (4 - 1).
2. Do the same with the real numbers. Multiplying 220122^{2012} by 22=42^2 = 4 adds 22 to the exponent, so
22014=4×22012.2^{2014} = 4 \times 2^{2012}.
3. Pull out the common part. Both the top and the bottom contain 220122^{2012}:
4×22012+220124×22012−22012=22012×(4+1)22012×(4−1).\frac{4 \times 2^{2012} + 2^{2012}}{4 \times 2^{2012} - 2^{2012}} = \frac{2^{2012} \times (4 + 1)}{2^{2012} \times (4 - 1)}.
4. Cancel. Now 220122^{2012} is multiplied on top and bottom, so it cancels, leaving 53\dfrac{5}{3}.

The answer is (C) 53\boxed{\textbf{(C)}\ \frac{5}{3}}.

Why this works

When you add or subtract powers of the same number, pull out the smallest power first. What remains is small and easy. You may only cancel something from top and bottom when it multiplies everything, which is why Step 3 comes before Step 4. Trying a small case (Step 1) is a good way to see the pattern and to check the answer.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Write 22014=22012⋅22=4⋅220122^{2014} = 2^{2012} \cdot 2^2 = 4 \cdot 2^{2012}. Then both the numerator and the denominator have the common factor 220122^{2012}:
22014+2201222014−22012=22012(4+1)22012(4−1)=53.\frac{2^{2014} + 2^{2012}}{2^{2014} - 2^{2012}} = \frac{2^{2012}(4 + 1)}{2^{2012}(4 - 1)} = \frac{5}{3}.

The answer is (C) 53\boxed{\textbf{(C)}\ \frac{5}{3}}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.