A twist on a classic

Art of Problem Solving

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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#1 h Mar 17, 2025, 3:01 PM

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Rank from smallest to largest: $\sqrt[2]{2}$ , $\sqrt[3]{3}$ , and $\sqrt[5]{5}$ .

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This post has been edited 4 times. Last edited by happypi31415, Mar 17, 2025, 3:06 PM

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#2 h Mar 17, 2025, 3:09 PM

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I finished all of the outside the box lol

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#3 h Mar 17, 2025, 5:29 PM

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put all of these to the 30th power: 2^15, 3^10, and 5^6. We have 3^2>2^3 so 3^10>2^15. also, 2^5>5^2, so 2^15>5^6. Thus, the order is $\boxed{\sqrt[5]{5},\sqrt[2]{2},\sqrt[3]{3}}$

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#4 h Mar 17, 2025, 11:56 PM

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happypi31415 wrote:

Rank from smallest to largest: $\sqrt[2]{2}$ , $\sqrt[3]{3}$ , and $\sqrt[5]{5}$ .

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$\sqrt[n]{n}<\sqrt[n-1]{n-1}$ for $n\geq2$ .

This post has been edited 1 time. Last edited by EaZ_Shadow, Mar 17, 2025, 11:56 PM

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#5 h Mar 18, 2025, 12:30 AM

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<- outside the box
there was a problem like this on high school math!
the one with 5 < one with 2 < one with 3 because 2 is weird.

This post has been edited 1 time. Last edited by krish6_9, Mar 18, 2025, 12:32 AM

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#6 h Mar 18, 2025, 12:37 AM

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By basic derivatives, the function $f(x)=x^{\frac{1}{x}}$ changes from being increasing to decreasing at $x=e.$ Thus, it is true that $f(n+1)<f(n)$ for $n>2.$

This post has been edited 1 time. Last edited by BS2012, Mar 18, 2025, 12:38 AM

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#7 h Mar 18, 2025, 12:44 AM

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Take to the $15$ th power. We get
$3^5 = 243$
$5^3 = 125$
$125<128< 2^\frac{15}{2} \approx \sqrt{2^{10} \cdot 2^5} \approx \sqrt{1000 \cdot 32} \approx \sqrt{32000} < \sqrt{32400}= 180<243$

This post has been edited 5 times. Last edited by Amkan2022, Mar 18, 2025, 12:47 AM

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#8 h Mar 18, 2025, 7:15 AM

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happypi31415 wrote:

Rank from smallest to largest: $\sqrt[2]{2}$ , $\sqrt[3]{3}$ , and $\sqrt[5]{5}$ .

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I think the optimal solution is to write the numbers in fractional powers, then raise each to the power of 30 (because that's the common denominator), and then compare natural numbers! Am I right?

This post has been edited 1 time. Last edited by invisibleman, Mar 18, 2025, 11:35 AM
Reason: typo

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#9 h Mar 18, 2025, 1:49 PM

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invisibleman wrote:

happypi31415 wrote:

Rank from smallest to largest: $\sqrt[2]{2}$ , $\sqrt[3]{3}$ , and $\sqrt[5]{5}$ .

Click to reveal hidden text

I think the optimal solution is to write the numbers in fractional powers, then raise each to the power of 30 (because that's the common denominator), and then compare natural numbers! Am I right?

Very solid. You do get 32768, 59049, and 15625 in which case the answer is 5thrt(5), sqrt(2), and cbrt(3), but in competition this would be difficult to multiply out. If you try raising to the 15th you get 128sqrt(2) (keep in mind that the square root of 2 is about 1.4), 243, and 125. You can rank it the same way, because simple scanning tells that 128*1.4 < 243. It's much easier to multiply out if you don't already know perfect powers of small numbers.

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#10 h Mar 19, 2025, 5:30 AM

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If you want to solve the problem without lengthy calculations, you can do so by working with mathematical analysis.
Consider the function f(x)=lnx/x, whose derivative is negative for x greather than e , so the function is decreasing. Now you can give values and compare the expressions.

This post has been edited 3 times. Last edited by invisibleman, Mar 19, 2025, 5:32 AM

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