Difference between revisions of "1987 AIME Problems/Problem 11"
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Thus <math>m(2n + m + 1) = 2 \cdot 3^{11}</math> so <math>m</math> is a [[divisor]] of <math>2\cdot 3^{11}</math>. However, because <math>n \geq 0</math> we have <math>m^2 < m(m + 1) \leq 2\cdot 3^{11}</math> so <math>m < \sqrt{2\cdot 3^{11}} < 3^6</math>. Thus, we are looking for large factors of <math>2\cdot 3^{11}</math> which are less than <math>3^6</math>. The largest such factor is clearly <math>2\cdot 3^5 = 486</math>; for this value of <math>m</math> we do indeed have the valid [[expression]] <math>3^{11} = 122 + 123 + \ldots + 607</math>, for which <math>k=\boxed{486}</math>. | Thus <math>m(2n + m + 1) = 2 \cdot 3^{11}</math> so <math>m</math> is a [[divisor]] of <math>2\cdot 3^{11}</math>. However, because <math>n \geq 0</math> we have <math>m^2 < m(m + 1) \leq 2\cdot 3^{11}</math> so <math>m < \sqrt{2\cdot 3^{11}} < 3^6</math>. Thus, we are looking for large factors of <math>2\cdot 3^{11}</math> which are less than <math>3^6</math>. The largest such factor is clearly <math>2\cdot 3^5 = 486</math>; for this value of <math>m</math> we do indeed have the valid [[expression]] <math>3^{11} = 122 + 123 + \ldots + 607</math>, for which <math>k=\boxed{486}</math>. | ||
− | == Solution 2== | + | === Solution 2=== |
First note that if <math>k</math> is odd, and <math>n</math> is the middle term, the sum is equal to kn. If <math>k</math> is even, then we have the sum equal to <math>kn+k/2</math> which is going to be even. Since <math>3^{11}</math> is odd, we see that <math>k</math> is odd. | First note that if <math>k</math> is odd, and <math>n</math> is the middle term, the sum is equal to kn. If <math>k</math> is even, then we have the sum equal to <math>kn+k/2</math> which is going to be even. Since <math>3^{11}</math> is odd, we see that <math>k</math> is odd. | ||
Latest revision as of 01:10, 13 May 2020
Problem
Find the largest possible value of for which is expressible as the sum of consecutive positive integers.
Solutions
Solution 1
Let us write down one such sum, with terms and first term :
.
Thus so is a divisor of . However, because we have so . Thus, we are looking for large factors of which are less than . The largest such factor is clearly ; for this value of we do indeed have the valid expression , for which .
Solution 2
First note that if is odd, and is the middle term, the sum is equal to kn. If is even, then we have the sum equal to which is going to be even. Since is odd, we see that is odd.
Thus, we have . Also, note Subsituting , we have . Proceed as in solution 1.
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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