Difference between revisions of "1987 AIME Problems/Problem 11"
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− | 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 | + | 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 will be even. Since <math>3^{11}</math> is odd, we see that <math>k</math> is odd. |
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. | 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. |
Revision as of 15:02, 3 January 2024
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 will 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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