Difference between revisions of "Talk:2010 AMC 12B Problems/Problem 25"
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f(x) = 78, 140, 162, 189, 164, 177, 162, 162, 149, 130, 128, 113, 108, 105, 99, 92, 83, 82, 77 | 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 <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 <math>\phantom{7*11*13=1001??????????}</math>will give an incorrect answer of <math>177</math>. The correct answer will be <math>164</math>. |
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? | ||
Revision as of 11:52, 18 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".
