2009 AIME II Problems/Problem 7
Define to be for odd and for even. When is expressed as a fraction in lowest terms, its denominator is with odd. Find .
First, note that , and that .
We can now take the fraction and multiply both the numerator and the denominator by . We get that this fraction is equal to .
Now we can recognize that is simply , hence this fraction is , and our sum turns into .
Let . Obviously is an integer, and can be written as . Hence if is expressed as a fraction in lowest terms, its denominator will be of the form for some .
In other words, we just showed that . To determine , we need to determine the largest power of that divides .
Let be the largest such that that divides .
We can now return to the observation that . Together with the obvious fact that is odd, we get that .
It immediately follows that , and hence .
Obviously, for the function is is a strictly decreasing function. Therefore .
We can now compute . Hence .
And thus we have , and the answer is .
Additionally, once you count the number of factors of in the summation, one can consider the fact that, since must be odd, it has to take on a value of or (Because the number of s in the summation is clearly greater than , dividing by will yield a number greater than , and multiplying this number by any odd number greater than will yield an answer , which cannot happen on the AIME.) Once you calculate the value of , and divide by , must be equal to , as any other value of will result in an answer . This gives as the answer.
Just a small note. It's important to note the properties of the function, which is what Solution 1 is using but denoting it as . We want to calculate as the final step. We know that one property of is that . Therefore, we have that . Thus, we see by similar calculations as in Solution 1, that . From which the conclusion follows.
Using the steps of the previous solution we get and if you do the small cases(like ) you realize that you can "thin-slice" the problem and simply look at the cases where (they're nearly identical in nature but one has with it) since hardly contains any powers of or in other words it's very inefficient and the inefficient cases hold all the power so you can just look at the highest powers of in and and you get the minimum power of in either expression is so the answer is since it would violate the rules of the AIME and the small cases if .
Solution 3 (Divisibility)
We can logically deduce that the value will be 1. Listing out the first few values of odd and even integers, we have: and . Obviously, none of the factors of in the denominator will cancel out, since the numerator is odd. Starting on the second term of the numerator, a factor of occurs every terms, and starting out on the third term of the denominator, a factor of appears also every terms. Thus, the factors of on the denominator will always cancel out. We can apply the same logic for every other odd factor, so once terms all cancel out, the denominator of the final expression will be in the form . Since there will be no odd factors in the denominators, all the denominators will be in the form where is the number of factors of in . This is simply . Therefore, our answer is .
Using the initial steps from Solution 1, . Clearly as in all the summands there are no non-power of 2 factors in the denominator. So we seek to find . Note that would be the largest denominator in all the summands, so when they are summed it is the common denominator.
Taking the p-adic valuations of each term, the powers of 2 in the denominator for is We can use Kummers theorem to see that is the number of digits carried over when is added to in base . This is simply the number of 's in the binary representation of .
Looking at the binary representations of some of the larger we see having eight 's. So the power of two is . Experimenting with we see that the power of two are all , and under the power of two . Therefore and
|2009 AIME II (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.