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Bunuel
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11^9 is 121 times greater than 9^9, which is already much smaller than 11^9. Straight away D
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Bunuel
Which of the following is the best approximation for \(11^{11}-9^9\)?

A. \(11^8\)
B. \(11^9\)
C. \(11^{10}\)
D. \(11^{11}\)
E. \(11^{12}\)




As explained by everyone above, \(9^9\) is much smaller than \(11^{11}\) and that is why subtracting \(9^9\) doesn't have much impact on the value of \(11^{11}\).

And if I had to prove it in this question, I would say, ok let me find out whether \(11^{11}-9^9\) is closer to \(11^{11}\) or \(11^{10}\) (the next smaller option)

\(11^{11} = 11 * 11^{10} = 121 * 11^9\)

So the ratio of \(11^9 : 11^{10} : 11^{11} = 1 : 11 : 121\)
Now if I remove 1 from 121, is the result 120 closer to 11 or 121? Obviously 121.
So if I remove \(11^9\) from \(11^{11}\), the result will be much closer to \(11^{11}\) than to \(11^{10}\).
Then if I were to remove an even smaller number \(9^9\) from \(11^{11}\), the result will obviously be much closer to \(11^{11}\) than to \(11^{10}\).

Hence the answer is \(11^{11}\) i.e. (D)
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Why is (9^9) treated as “tiny” compared to (11^11)?

Core idea:
Don’t think “Is 9^9 big?”
Think → “Is 9^9 big relative to 11^11?”

Step 1 — Compare using ratio

(11^11) / (9^9) = (11/9)^9 × 11^2

Approximate:
11/9 ≈ 1.2
1.2^9 ≈ 5
11^2 = 121

So:
ratio ≈ 5 × 121 ≈ 600

Step 2 — GMAT intuition

Subtracting a very small fraction:
600 − 1 ≈ 600

So:
11^11 − 9^9 ≈ 11^11

Key Rule (store this):
If one term is hundreds of times smaller, subtraction ≈ no change

Contrast (important):

Cannot ignore:
11^11 − 10^11
Ratio ≈ 3 → not small
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