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

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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 1

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

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

Suppose instead we take an even number $2k$ of consecutive integers, with median being the average of $m$ 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 $1,3,5$ with medians $(172,173),(57,58),(34,35)$ respectively.

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

Solution 2

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: $\frac {n \cdot (k+g)}{2}$ .

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 substiute our new value of $g$ into $\frac {n \cdot (k+g)}{2}$ to get that the sum is $\frac {n \cdot (k+k+n-1)}{2} = 345$ .

This simplifies to $\frac {n \cdot (2k+n-1)}{2} = 345$ . This gives a nice equation. We multiply out the 2 to get that $n \cdot (2k+n-1)=690$ . 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 $\boxed{\textbf{(E) }7}$ ways.

~~jk23541

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 $30$ , and that is because $\frac{345}{30}=11.5$ , which means $11$ and $12$ must be the middle 2 numbers; however, a sequence of length $30$ with middle numbers $11$ and $12$ that consists only of integers would go into the negatives, so any number from 30 onwards wouldn't work, and since $1$ is a trivial, non-counted solution, we get $\boxed{\textbf{(E) }7}$ -ColtsFan10

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 ==See Also==
 {{AMC10 box|year=2016|ab=B|num-b=17|num-a=19}}
+[[Category:Introductory Number Theory Problems]]
 {{MAA Notice}}

2016 AMC 10B (Problems • Answer Key • Resources)
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