2016 AMC 10B Problems/Problem 18
Problem
In how many ways can 
be written as the sum of an increasing sequence of two or more consecutive positive integers?
Solution 1
Factor 
.
Suppose we take an odd number 
of consecutive integers, with the median as 
. Then 
with 
.
Looking at the factors of , the possible values of are 
with medians as 
respectively.
Suppose instead we take an even number 
of consecutive integers, with median being the average of and 
. Then 
with 
.
Looking again at the factors of , the possible values of are 
with medians 
respectively.
Thus the answer is 
.
Solution 2
We need to find consecutive numbers (an arithmetic sequence that increases by 
) that sums to . This calls for the sum of an arithmetic sequence given that the first term is , the last term is 
and with 
elements, which is: 
.
So, since it is a sequence of consecutive numbers starting at and ending at 
. We can now substitute with . Now we substiute our new value of into to get that the sum is 
.
This simplifies to 
. This gives a nice equation. We multiply out the 2 to get that 
. This leaves us with 2 integers that multiplies to 
which leads us to think of factors of . We know the factors of are: 
. So through inspection (checking), we see that only 
and 
work. This gives us the answer of ways.
~~jk23541
An alternate way to finish.
Let
\begin{align*}
2k+n-1 &=\frac{690}{k} \\
n &=\frac{k} \\
\end{align*} (Error compiling LaTeX. Unknown error_msg)
where is a factor of 
We find 
so we need 
to be positive and odd. Fortunately, regardless of the parity of we see that is odd. Furthermore, we need 
which eliminates exact half of the factors. Now, since we need more than integer to sum up we need 
which eliminates one more case. There were 
cases to begin with, so our answer is 
ways.
Solution 2.1
At the very end of Solution 2, where we find the factors of 690, instead of inspection, notice that all numbers will work until you get to 
, and that is because 
, which means 
and 
must be the middle 2 numbers; however, a sequence of length with middle numbers and that consists only of integers would go into the negatives, so any number from 30 onwards wouldn't work, and since is a trivial, non-counted solution, we get -ColtsFan10
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. 
