Difference between revisions of "2016 AMC 10B Problems/Problem 18"
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Factorize <math>345=3\cdot 5\cdot 23</math>. | Factorize <math>345=3\cdot 5\cdot 23</math>. | ||
| − | Suppose we take an odd number <math>k</math> of consecutive integers, | + | Suppose we take an odd number <math>k</math> of consecutive integers, centered on <math>m</math>. Then <math>mk=345</math> with <math>\tfrac12k<m</math>. |
Looking at the factors of <math>345</math>, the possible values of <math>k</math> are <math>3,5,15,23</math> centred on <math>115,69,23,15</math> respectively. | Looking at the factors of <math>345</math>, the possible values of <math>k</math> are <math>3,5,15,23</math> centred on <math>115,69,23,15</math> respectively. | ||
Revision as of 14:22, 18 June 2016
Problem
In how many ways can 
be written as the sum of an increasing sequence of two or more consecutive positive integers?
Solution
Factorize 
.
Suppose we take an odd number 
of consecutive integers, centered on 
. Then 
with 
.
Looking at the factors of , the possible values of are 
centred on 
respectively.
Suppose instead we take an even number 
of consecutive integers, centred on and 
. Then 
with 
.
Looking again at the factors of , the possible values of are 
centred on 
respectively.
Thus the answer is 
.
See Also
| 2016 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
