Difference between revisions of "2016 AMC 10B Problems/Problem 18"
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| − | We | + | We need to find consecutive numbers (an arithmetic sequence that increases by <math>1</math>) that sums to <math>345</math>. This calls for the sum of an arithmetic sequence given that the first term is <math>k</math>, the last term is <math>g</math> and with <math>n</math> elements, which is: <math>\frac {n*(k+g)}{2}</math>. |
So since it is a sequence of <math>n</math> consecutive numbers starting at <math>k</math> and ending at <math>k+n-1</math>. We can now substitute <math>g</math> with <math>k+n-1</math>. Now we subsittute our new value of <math>g</math> into <math>\frac {n*(k+g)}{2}</math> to get that the sum is <math>\frac {n*(k+k+n-1)}{2} = 345</math>. | So since it is a sequence of <math>n</math> consecutive numbers starting at <math>k</math> and ending at <math>k+n-1</math>. We can now substitute <math>g</math> with <math>k+n-1</math>. Now we subsittute our new value of <math>g</math> into <math>\frac {n*(k+g)}{2}</math> to get that the sum is <math>\frac {n*(k+k+n-1)}{2} = 345</math>. | ||
Revision as of 16:07, 2 January 2018
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: 
.
So since it is a sequence of consecutive numbers starting at and ending at 
. We can now substitute with . Now we subsittute 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
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. 
