Difference between revisions of "2016 AMC 10B Problems/Problem 18"
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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>. | 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 | + | 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 substiute 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>. |
This simplifies to <math>\frac {n*(2k+n-1)}{2} = 345</math>. This gives a nice equation. We multiply out the 2 to get that <math>n*(2k+n-1)=690</math>. This leaves us with 2 integers that multiplies to <math>690</math> which leads us to think of factors of <math>690</math>. We know the factors of <math>690</math> are: <math>1,2,3,5,6,10,15,23,30,46,69,115,138,230,345,690</math>. So through inspection (checking), we see that only <math>2,3,5,6,10,15</math> and <math>23</math> work. This gives us the answer of <math>\textbf{(E) }7</math> ways. | This simplifies to <math>\frac {n*(2k+n-1)}{2} = 345</math>. This gives a nice equation. We multiply out the 2 to get that <math>n*(2k+n-1)=690</math>. This leaves us with 2 integers that multiplies to <math>690</math> which leads us to think of factors of <math>690</math>. We know the factors of <math>690</math> are: <math>1,2,3,5,6,10,15,23,30,46,69,115,138,230,345,690</math>. So through inspection (checking), we see that only <math>2,3,5,6,10,15</math> and <math>23</math> work. This gives us the answer of <math>\textbf{(E) }7</math> ways. |
Revision as of 00:32, 21 November 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 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.
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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 |
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