Talk:2010 AMC 12B Problems/Problem 25

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 $164$ .

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

--- buhiroshi0205

It just so happens that this competition is a 75-minute, 25-problem multiple-choice contest, not a 4.5-hour-per-day, 2-day, 6-question olympiad. In these types of contests, with three minutes to solve each problem and computational aids (like calculators, smartwatches, abaci, and your beloved Wolfram Mathematica), by the time you get to #25 you're already pressed for time. The type of unnecessary rigor you're asking for here is tantamount to finding $v_2(n!)$ and $v_5(n!)$ every time you're asked to find the number of trailing zeros in a factorial—sure, go ahead, but you're just wasting your own time. And the problem asked for $2010^m,$ not $2001^m.$ In this case, $67$ works, so it works. No need of proof by Wolfram Mathematica or two-paragraph-long AoPSWiki edit. If you're so interested in the fact that $f(x)$ is not strictly decreasing, you may have stumbled on some great revelation, but such revelations are not suited for the time-crunched, half-BS'ed solutions to AMC problems. If this fact still piques your interest, try submitting your problem to MITComPOsiTeS or RSJ or some other research program. I'm sure they'd be happy to help you.

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Talk:2010 AMC 12B Problems/Problem 25