Difference between revisions of "2016 AMC 12B Problems/Problem 16"
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==Solution== | ==Solution== | ||
| + | We proceed with this problem by considering two cases, when: \par | ||
| + | 1. There are an odd number of consecutive numbers \par | ||
| + | 2. There are an even number of consecutive numbers | ||
| + | |||
| + | For the first case, we can cleverly choose the convenient form of our sequence to be | ||
| + | <cmath>a-n,\cdots, a-1, a, a+1, \cdots, a+n</cmath> | ||
| + | |||
| + | because then our sum will just be <math>(2n+1)a</math>. We now have | ||
| + | <cmath>(2n+1)a = 345</cmath> | ||
| + | and <math>a</math> will have a solution when <math>\frac{345}{2n+1}</math> is an integer, namely when <math>2n+1</math> is a divisor of 345. We check that | ||
| + | <cmath>2n+1 = 3, 5, 15, 23</cmath> | ||
| + | work, and no more, because <math>2n+1</math> does not satisfy the requirements of two or more consecutive integers, and when <math>2n+1</math> equals the next biggest factor, <math>69</math>, there must be negative integers in the sequence. Our solutions are <math>\{114,115, 116\}, \{67, \cdots, 71\}, \{16, \cdots, 30\}, \{4, \cdots, 26\}</math>. | ||
| + | |||
| + | For the even cases, we choose our sequence to be of the form: | ||
| + | <cmath>a-(n-1), \cdots, a, a+1, \cdots, a+n</cmath> | ||
| + | so the sum is <math>\frac{(2n)(2a+1)}{2} = n(2a+1)</math>. In this case, we find our solutions to be <math>\{172, 173\}, \{55,\cdots, 60\}, \{30,\cdots, 39\}</math>. | ||
| + | |||
| + | We have found all 7 solutions and our answer is <math>\boxed{\textbf{(E)} \, 7}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2016|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:44, 21 February 2016
Problem
In how many ways can 
be written as the sum of an increasing sequence of two or more consecutive positive integers?
Solution
We proceed with this problem by considering two cases, when: \par 1. There are an odd number of consecutive numbers \par 2. There are an even number of consecutive numbers
For the first case, we can cleverly choose the convenient form of our sequence to be
![\[a-n,\cdots, a-1, a, a+1, \cdots, a+n\]](http://latex.artofproblemsolving.com/7/a/0/7a04526e358b74f6c02e0d6197b3ccfda2b03a8e.png)
because then our sum will just be 
. We now have
![\[(2n+1)a = 345\]](http://latex.artofproblemsolving.com/3/1/1/31121d247e7bb15216675853125c76f8f7916c9d.png)
and 
will have a solution when 
is an integer, namely when 
is a divisor of 345. We check that
![\[2n+1 = 3, 5, 15, 23\]](http://latex.artofproblemsolving.com/4/6/e/46e8888ffd22a53e9d54b886173734fb9e0dffda.png)
work, and no more, because does not satisfy the requirements of two or more consecutive integers, and when equals the next biggest factor, 
, there must be negative integers in the sequence. Our solutions are 
.
For the even cases, we choose our sequence to be of the form:
![\[a-(n-1), \cdots, a, a+1, \cdots, a+n\]](http://latex.artofproblemsolving.com/d/d/c/ddc4567a8fec244e7d8b199e289a10b2a5730020.png)
so the sum is 
. In this case, we find our solutions to be 
.
We have found all 7 solutions and our answer is 
.
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
