Difference between revisions of "2009 AMC 12A Problems/Problem 18"
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For <math>k\in\{1,2,3\}</math> we have <math>I_k = 2^{k+2} \left( 5^{k+2} + 2^{k-4} \right)</math>. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have <math>N(k)=k+2\leq 5</math>. | For <math>k\in\{1,2,3\}</math> we have <math>I_k = 2^{k+2} \left( 5^{k+2} + 2^{k-4} \right)</math>. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have <math>N(k)=k+2\leq 5</math>. | ||
− | For <math>k> | + | For <math>k>4</math> we have <math>I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)</math>. For <math>k>4</math> the value in the parentheses is odd, hence <math>N(k)=6</math>. |
This leaves the case <math>k=4</math>. We have <math>I_4 = 2^6 \left( 5^6 + 1 \right)</math>. The value <math>5^6 + 1</math> is obviously even. And as <math>5\equiv 1 \pmod 4</math>, we have <math>5^6 \equiv 1 \pmod 4</math>, and therefore <math>5^6 + 1 \equiv 2 \pmod 4</math>. Hence the largest power of <math>2</math> that divides <math>5^6+1</math> is <math>2^1</math>, and this gives us the desired maximum of the function <math>N</math>: <math>N(4) = \boxed{7}</math>. | This leaves the case <math>k=4</math>. We have <math>I_4 = 2^6 \left( 5^6 + 1 \right)</math>. The value <math>5^6 + 1</math> is obviously even. And as <math>5\equiv 1 \pmod 4</math>, we have <math>5^6 \equiv 1 \pmod 4</math>, and therefore <math>5^6 + 1 \equiv 2 \pmod 4</math>. Hence the largest power of <math>2</math> that divides <math>5^6+1</math> is <math>2^1</math>, and this gives us the desired maximum of the function <math>N</math>: <math>N(4) = \boxed{7}</math>. |
Revision as of 16:13, 2 October 2016
- The following problem is from both the 2009 AMC 12A #18 and 2009 AMC 10A #25, so both problems redirect to this page.
Contents
[hide]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.