Difference between revisions of "2009 AMC 12A Problems/Problem 18"
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Expilncalc (talk | contribs) m (→Alternate Solution: Edit: The EDIT beforehand has been resolved, so I put that in quotes. Meanwhile, I found out something else.) |
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Observe then that <math>\boxed{\textbf{(B)}7}</math> must be the maximum value for <math>N(k)</math> because whatever value we choose for <math>k</math>, <math>N(k)</math> must be less than or equal to <math>7</math>. | Observe then that <math>\boxed{\textbf{(B)}7}</math> must be the maximum value for <math>N(k)</math> because whatever value we choose for <math>k</math>, <math>N(k)</math> must be less than or equal to <math>7</math>. | ||
− | + | "Isn't this solution incomplete because we need to show that <math>N(k) = 7</math> can be reached?" | |
An example of 7 being reached is 1000064. 1000064 divided by <math>2^7=128</math> is 7813. | An example of 7 being reached is 1000064. 1000064 divided by <math>2^7=128</math> is 7813. | ||
+ | |||
+ | In fact, 1000064 is the ONLY <math>N(k)</math> that satisfies 7. All others are 6 or lower, because if there are more zeroes, to be divisible by 128 ($2^7), the last 7 digits must be divisible by 128, but 64 isn't. Meanwhile, if there are less zeroes, we can test by division that they do not work. | ||
== See Also == | == See Also == |
Revision as of 19:02, 21 January 2017
- The following problem is from both the 2009 AMC 12A #18 and 2009 AMC 10A #25, so both problems redirect to this page.
Problem
For , let , where there are zeros between the and the . Let be the number of factors of in the prime factorization of . What is the maximum value of ?
Solution
The number can be written as .
For we have . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have .
For we have . For the value in the parentheses is odd, hence .
This leaves the case . We have . The value is obviously even. And as , we have , and therefore . Hence the largest power of that divides is , and this gives us the desired maximum of the function : .
Alternate Solution
Notice that 2 is a prime factor of an integer if and only if is even. Therefore, given any sufficiently high positive integral value of , dividing by yields a terminal digit of zero, and dividing by 2 again leaves us with where is an odd integer. Observe then that must be the maximum value for because whatever value we choose for , must be less than or equal to .
"Isn't this solution incomplete because we need to show that can be reached?"
An example of 7 being reached is 1000064. 1000064 divided by is 7813.
In fact, 1000064 is the ONLY that satisfies 7. All others are 6 or lower, because if there are more zeroes, to be divisible by 128 ($2^7), the last 7 digits must be divisible by 128, but 64 isn't. Meanwhile, if there are less zeroes, we can test by division that they do not work.
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.