2016 AMC 10B Problems/Problem 18
Contents
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: .
We look for sequences 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 multiply 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 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
Solution 3 (Fast And Clean)
The median of the sequence is either an integer or a half integer. Let , then .
On the other hand we have two constraints:
1) because the integers in the sequence are all positive, and ;
2) If is odd then is an integer, is even; if is even then is a half integer, is odd. Therefore, and have opposite parity.
Now has factors and it is not a perfect square. There are choices for . Also since , we know and must have opposite parity. Therefore the answer is .
~ asops
Video Solution 1
~savannahsolver
Video Solution 2
https://youtu.be/ZhAZ1oPe5Ds?t=950
~ pi_is_3.14
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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All AMC 10 Problems and Solutions |
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