2006 AIME I Problems/Problem 13
Contents
Problem
For each even positive integer , let denote the greatest power of 2 that divides For example, and For each positive integer let Find the greatest integer less than 1000 such that is a perfect square.
Solution 1
Given , consider . Define . There are elements of that are divisible by , elements of that are divisible by but not by and elements of that are divisible by but not by .
Thus Let be the highest power of that divides . Thus by the above formula, the highest power of that divides is . For to be a perfect square, must be even. If is odd, then is even, hence is odd, and cannot be a perfect square. Hence must be even. In particular, as , we have five choices for , namely .
If , then is odd, so is odd, hence the largest power of dividing has an odd exponent, so is not a perfect square.
In the other cases, note that is even, so the highest power of dividing will be a perfect square. In particular, will be a perfect square if and only if is an odd perfect square.
If , then implies that , so we have .
If , then implies that , so .
If , then implies that , so .
If , then implies that , so .
Comparing the largest term in each case, we find that the maximum possible such that is a perfect square is .
Solution 2
First note that if is odd and if is even. so must be odd so this reduces to Thus Further noting that we can see that which is the same as above. To simplify the process of finding the largest square we can note that if is odd then must be exactly divisible by an odd power of . However, this means is even but it cannot be. Thus is even and is a large even square. The largest even square is so
Solution 3 (Finding patterns and using recursion)
At first, this problem looks kind of daunting, but we can easily solve this problem by finding patterns, recursion and algebraic manipulations.
We first simplify all the messy notation in the term. Note that the problem asks us to find the smallest value of such that there exists an integer that satisfies
.
Since there is no obvious way to approach this problem, we start by experimenting with small values of to evaluate some .
We play with these values:
We are certainly not going to expand all of this out... so let's look for patterns from these values!
Using a little bit of ingenuity, we note that
Aha! We see powers of two in each of our terms! Therefore, we can say that
Hooray! We have a recursion! Realistically, we would want to prove that the recursion works, but I currently don't know how to prove it.
On the actual AIME, go with whatever patterns you see, because most likely those are the patterns that the test-makers want the students to see.
So we may generalize a formula for :
Uh oh... this formula is not in closed form. Looks like we'll have to use our recursion to develop one manually. We do so by using our recursion for :
Let's simplify a bit further, where we use our recursion for .
We now see a pattern! Using the exact same logic, we can condense this whole messy thing into a closed form:
We have our closed form, so now we can find the largest value of such that is a perfect square.
In order for to be a perfect square, we must have even and be a perfect square.
Since , we have . We first try (since it is the smallest square below , which gives us . But isn't even, so we discard this value.
Next, we try the second smallest value, which is , which tells us that . is indeed even, and is a perfect square, so the largest value of such that is a perfect square is .
Our answer is .
-FIREDRAGONMATH16
Solution 4
First, we set intervals. Say that a number falls strictly within .
We can generalize this:
If a number is in the form where is a positive odd number, :
so there are numbers that satisfy this form. For all numbers that satisfy this form, . The sum of for all in this form and interval is . can vary between and , so the total sum is . The domain of the function we are trying to find is all even integers in the interval so there are values of . Now we have the sum . However, we did not consider powers of two yet(since our interval was strictly between powers of 2), so we have to add . Note that . Adding , we get . If this sum is a perfect square, , since that would make not a perfect square, and odd so it cannot contribute a factor of 2 to make the power of 2 a perfect square.
We want the least odd number less than such that is an even perfect square. The greatest even square less than is so
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.