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Difference between revisions of "2016 AMC 10B Problems/Problem 18"

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We have that 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 have that 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>, the sum would be <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>.
  
 
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 19:06, 16 January 2017

Problem

In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution

Factorize $345=3\cdot 5\cdot 23$.

Suppose we take an odd number $k$ of consecutive integers, centered on $m$. Then $mk=345$ with $\tfrac12k<m$. Looking at the factors of $345$, the possible values of $k$ are $3,5,15,23$ centred on $115,69,23,15$ respectively.

Suppose instead we take an even number $2k$ of consecutive integers, centred on $m$ and $m+1$. Then $k(2m+1)=345$ with $k\le m$. Looking again at the factors of $345$, the possible values of $k$ are $1,3,5$ centred on $(172,173),(57,58),(34,35)$ respectively.

Thus the answer is $\textbf{(E) }7$.


Solution 2

We have that we need to find consecutive numbers (an arithmetic sequence that increases by $1$) that sums to $345$. This calls for the sum of an arithmetic sequence given that the first term is $k$, the last term is $g$ and with $n$ elements, which is: $\frac {n*(k+g)}{2}$.

So since it is a sequence of $n$ consecutive numbers starting at $k$ and ending at $k+n-1$. We can now substitute $g$ with $k+n-1$. Now we subsittute our new value of $g$ into $\frac {n*(k+g)}{2}$ to get that the sum is $\frac {n*(k+k+n-1)}{2} = 345$.

This simplifies to $\frac {n*(2k+n-1)}{2} = 345$. This gives a nice equation. We multiply out the 2 to get that $n*(2k+n-1)=690$. This leaves us with 2 integers that multiplies to $690$ which leads us to think of factors of $690$. We know the factors of $690$ are: $1,2,3,5,6,10,15,23,30,46,69,115,138,230,345,690$. So through inspection (checking), we see that only $2,3,5,6,10,15$ and $23$ work. This gives us the answer of $\textbf{(E) }7$ 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

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