Difference between revisions of "1987 AIME Problems/Problem 11"
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== Problem == | == Problem == | ||
Find the largest possible value of <math>k</math> for which <math>3^{11}</math> is expressible as the sum of <math>k</math> consecutive [[positive integer]]s. | Find the largest possible value of <math>k</math> for which <math>3^{11}</math> is expressible as the sum of <math>k</math> consecutive [[positive integer]]s. | ||
− | == Solution == | + | == Solution 1== |
Let us write down one such sum, with <math>m</math> terms and first term <math>n + 1</math>: | Let us write down one such sum, with <math>m</math> terms and first term <math>n + 1</math>: | ||
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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>. | ||
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+ | == Solution 2== | ||
+ | First note that if k is odd, and n is the middle term, the sum is equal to kn. If k is even, then we have the sum equal to kn+k/2 which is going to be even. Since 3^11 is odd, we see that k is odd. | ||
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+ | Thus, we have <math>nk=3^{11} \implies n=3^{11}/k</math>. Also, note <math>n-(k+1)/2=0 \implies n=(k+1)/2.</math> Subsituting <math>n=3^{11}/k</math>, we have <math>k^2+k=2*3^{11}</math>. Proceed as in solution 1. | ||
== See also == | == See also == |
Revision as of 15:01, 3 February 2018
Contents
Problem
Find the largest possible value of for which is expressible as the sum of consecutive positive integers.
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 k is odd, and n is the middle term, the sum is equal to kn. If k is even, then we have the sum equal to kn+k/2 which is going to be even. Since 3^11 is odd, we see that k 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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