# 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 but , then 13 as the largest prime in 2001=7*11*13 will give an incorrect answer of . The correct answer will be .

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 and , 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