Difference between revisions of "2009 AMC 12A Problems/Problem 18"
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Notice that 2 is a prime factor of an integer <math>n</math> if and only if <math>n</math> is even. Therefore, given any sufficiently high positive integral value of <math>k</math>, dividing <math>I_k</math> by <math>2^6</math> yields a terminal digit of zero, and dividing by 2 again leaves us with <math>2^7 \cdot a = I_k</math> where <math>a</math> is an odd integer. | Notice that 2 is a prime factor of an integer <math>n</math> if and only if <math>n</math> is even. Therefore, given any sufficiently high positive integral value of <math>k</math>, dividing <math>I_k</math> by <math>2^6</math> yields a terminal digit of zero, and dividing by 2 again leaves us with <math>2^7 \cdot a = I_k</math> where <math>a</math> is an odd integer. | ||
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>. | ||
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+ | EDIT: Isn't this solution incomplete because we need to show that <math>N(k) = 7</math> can be reached? | ||
== See Also == | == See Also == |
Revision as of 11:12, 31 January 2016
- 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 .
EDIT: Isn't this solution incomplete because we need to show that can be reached?
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.