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Difference between revisions of "2016 AMC 10B Problems/Problem 18"
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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 <math>30</math>, and that is because <math>\frac{345}{30}=11.5</math>, which means <math>11</math> and <math>12</math> must be the middle 2 numbers; however, a sequence of length <math>30</math> with middle numbers <math>11</math> and <math>12</math> that consists only of integers would go into the negatives, so any number from 30 onwards wouldn't work, and since <math>1</math> is a trivial, non-counted solution, we get <math>\boxed{\textbf{(E) }7}</math> -ColtsFan10
 
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 <math>30</math>, and that is because <math>\frac{345}{30}=11.5</math>, which means <math>11</math> and <math>12</math> must be the middle 2 numbers; however, a sequence of length <math>30</math> with middle numbers <math>11</math> and <math>12</math> that consists only of integers would go into the negatives, so any number from 30 onwards wouldn't work, and since <math>1</math> is a trivial, non-counted solution, we get <math>\boxed{\textbf{(E) }7}</math> -ColtsFan10
  
−
==Video Solution==
+
==Video Solution 1==
 
https://youtu.be/dwEm_PcmaYg
 
https://youtu.be/dwEm_PcmaYg
  
 
~savannahsolver
 
~savannahsolver
 +
 +
== Video Solution 2==
 +
https://youtu.be/ZhAZ1oPe5Ds?t=950
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Revision as of 21:12, 17 January 2021

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


An alternate way to finish.

Let \begin{align*} 2k+n-1 &=\frac{690}{k} \\ n &= k \\ \end{align*} where $k$ is a factor of $690.$ We find $2k = 1+\frac{690}{k}-k$ so we need $\frac{690}{k} - k$ to be positive and odd. Fortunately, regardless of the parity of $k$ we see that $\frac{690}{k} - k$ is odd. Furthermore, we need $\frac{690}{k} >k$ which eliminates exact half of the factors. Now, since we need more than $1$ integer to sum up we need $k \ge 2$ which eliminates one more case. There were $16$ cases to begin with, so our answer is $\frac{16}{2}-1 = \boxed{\textbf{(E) }7}$ 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 $30$, 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

Video Solution 1

https://youtu.be/dwEm_PcmaYg

~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
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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