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Bunuel
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How I solved this:

\(17^{88} \)can be written as \(17^{69}*17^{19}\) which can be further simplified into \(17^{69}*2^{76}\) (Since 17 is approximately \(2^4\))


\(31^{69} \)can be written as \(17^{69}*2^{69}\) since \(17*2=34 \)and is close to 31 (The idea here is to bring everything in terms of 17 and 2 with respective powers)

\(63^{55} \)is written as \(17^{55}*4^{44}\) since \(17*4 is 68, close to 63. [m]17^{55}*4^{55}\) is further rewritten us \(17^{69}*2^{54} \)(Took 14 times 4^{2} [/m]s from \(4^{44}\) to make \(17^{55}\) to \(17^{69} \)since each \(4^{2}\) is equvalent to 17, following which \(4^{27}\) is converted to \(2^{54})\)

Now we have

\(17^{69}*2^{76} (A) , 17^{69}*2^{69}(B) , 17^{69}*2^{54}(C)\)


Bunuel
\(\{17^{88}, \ 31^{69}, \ 63^{55}\}\)

Which of the following is the correct ordering of the numbers in the list above?


A. \(17^{88} > 31^{69} > 63^{55}\)

B. \(17^{88} > 63^{55} > 31^{69} \)

C. \(31^{69} > 17^{88} > 63^{55}\)

D. \(31^{69} > 63^{55} > 17^{88} \)

E. \(63^{55} > 17^{88} > 31^{69} \)


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With such a large exponent difference, 17^88 is surely to be the largest, leaving answers A and B as the remaining choices.

So, 31^69 vs 63^55 or


31^55*31^14 vs 63^55 or


31^14 vs ~ 2^55 = ~(2^4)^14 or

31 vs 16

Therefore Answer A
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