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Among the five real numbers below, which one is the smallest?

$\text{(A)}\ \ \sqrt[\leftroot{-2}\uproot{2}2009]{2010};\quad\text{(B)}\ \ \sqrt[\leftroot{-2}\uproot{2}2010]{2009};\quad\text{(C)}\ \ 2010;\quad\text{(D)}\ \ \frac{2010}{2009};\quad\text{(E)}\ \ \frac{2009}{2010}.$

Among the five integers below, which one is the largest?

$\text{(A)}\ \ 2009^{2010};\quad\text{(B)}\ \ 20092010^2;\quad\text{(C)}\ \ 2010^{2009};\quad\text{(D)}\ \ {3}^{(3^{(3^{3})})};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \text{(E)}\ \ 2^{10}+4^{10}+\cdots+2010^{10}$

As you can see, the above are the questions from $SMO - 2010$, I want to know if there exists a uniform method to solve such kind of questions.

All the options look equally promising.

In the $1^{st}$ part, option $2010$, $\frac{2010}{2009}$ and $\frac{2009}{2010}$get eliminated. I have no idea about how to make comparison between $A$ and $B$.

Similarly in the $2^{nd}$ part, I am clueless.

Please tell a uniform method which applies for all questions. Thanks.

Bill Dubuque
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    $2009^{1/2010}<2010^{1/2009}$. In general, we have $n^{1/(n+1)}<(n+1)^{1/n}$. Search this here! – Dietrich Burde Mar 22 '21 at 12:03
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    In first part, E is the only option smaller than $1$. For second part, you would see Fastest way to check if $x^y>y^x$? (if both $x,y$ are larger than $e\approx 2.7183$ then you can just check $x \lt y$) to determine the answer between A and C, if D was not the largest one already (see Computation of Digits in Tetration). – Vepir Mar 22 '21 at 12:24
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    I think there is no "uniform method". You have to know some things about how numbers work, and how quickly certain functions increase compared with one another. You said of question 1, “All the options look equally promising.” Is this really true, and you don't see immediately that (C) is wrong? I want to write an answer, but I'm not sure what points you will find helpful. – MJD Mar 22 '21 at 12:34
  • @MJD Of course (C) gets rejected immediately, I was actually talking about comparison of (A) and (B). Please go ahead and write an answer, because I'm not quite sure about how to proceed. Although the points given by Dietrich and Vepir are helpful, I'm not sure if it helps in each case. Perhaps it requires a general intuition, which will come through practice. Nevertheless, please go ahead and share your answer. –  Mar 22 '21 at 16:32
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    For the first part, why is 2009/2010 eliminated? IMO You just have to recognize that term is <1, and the rest is > 1. – Calvin Lin Mar 22 '21 at 16:40

1 Answers1

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As mentioned in the comments, there isn't a "uniform" method. Some ideas are

  • Spot a pattern,
  • Compare to constants/powers like $1, 10^k, 2^k$.
  • Be aware of which functions grow quickly / slowly.

Hint for part 1: Show that one term is $<1$ and all the other terms are $>1$.
I'm guessing options A and B were put there to "scare" people / make them think the question is harder than it actually is.

Hint for part 2: The idea here is that tower of exponents grow very quickly. You should be aware of this fact, and so that is the most likely guess without further computation.

To prove it, you could bound 3^3^3^3 $> 10^{8040} > \max ( 2010^{2009}, 2009^{2010} )$.

If you're stuck, use $ 3^5 > 100$.

3^3^27 > $3^{10000000000} > 100^{10000000000/5 } $.

Calvin Lin
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    @Phymathechem. Thanks for wanting to make that edit. I intentionally left it that way, as I find it easier to read. Note that towers of exponents are always evaluated from the top down, so there is no ambiguity. – Calvin Lin Mar 22 '21 at 16:51