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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 chetan2u KarishmaB can you share an explanation for this question please?
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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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