Difference between revisions of "Talk:2010 AMC 12B Problems/Problem 25"

 
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So if the target problem was not <math>2010^m</math> but <math>2001^m</math>, then 13 as the largest prime in 2001=7*11*13 will give an incorrect answer of <math>177</math>. The correct answer will be <math>164</math>.
 
So if the target problem was not <math>2010^m</math> but <math>2001^m</math>, then 13 as the largest prime in 2001=7*11*13 will give an incorrect answer of <math>177</math>. The correct answer will be <math>164</math>.
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<math>\phantom{7*11*13=1001??????????}</math>
 
<math>\phantom{7*11*13=1001??????????}</math>
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Do we just have to cross our fingers and guess that it is f(67)=77 is indeed the smallest?
 
Do we just have to cross our fingers and guess that it is f(67)=77 is indeed the smallest?
  
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Because the AMC has a short time limit, yes, you "just have to cross our fingers and guess".
 
Because the AMC has a short time limit, yes, you "just have to cross our fingers and guess".
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Looking at the results of the Mathematica function mentioned above, you would likely want to guess both <math>2</math> and <math>67</math>, as the function appears to increase then decrease. That's probably the best way to have a chance of solving the problem quickly, and hope that they didn't throw in an outlier like 11.
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--- donutvan

Latest revision as of 11:54, 29 August 2021

How do we know that 67 will yield the smallest result of 77?

I created a Mathematica function to check all primes <= 67, and yes indeed 67 gave the smallest result of 77 but 2 gives a problematically close 78. And the trend is not always decreasing:

   x =  2,   3,   5,   7,  11,  13,  17,  19,  23,  29,  31,  37,  41,  43, 47, 53, 59, 61, 67  
f(x) = 78, 140, 162, 189, 164, 177, 162, 162, 149, 130, 128, 113, 108, 105, 99, 92, 83, 82, 77

So if the target problem was not $2010^m$ but $2001^m$, then 13 as the largest prime in 2001=7*11*13 will give an incorrect answer of $177$. The correct answer will be $164$.

$\phantom{7*11*13=1001??????????}$

Do we just have to cross our fingers and guess that it is f(67)=77 is indeed the smallest?

--- buhiroshi0205

Because the AMC has a short time limit, yes, you "just have to cross our fingers and guess".

Looking at the results of the Mathematica function mentioned above, you would likely want to guess both $2$ and $67$, as the function appears to increase then decrease. That's probably the best way to have a chance of solving the problem quickly, and hope that they didn't throw in an outlier like 11.

--- donutvan

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