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

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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7</math>
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7</math>
  
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==Dumb Solution==
+
==Solution 1==
 
Factor <math>345=3\cdot 5\cdot 23</math>.
 
Factor <math>345=3\cdot 5\cdot 23</math>.
  
Line 14: Line 14:
 
Looking again at the factors of <math>345</math>, the possible values of <math>k</math> are <math>1,3,5</math> with medians <math>(172,173),(57,58),(34,35)</math> respectively.
 
Looking again at the factors of <math>345</math>, the possible values of <math>k</math> are <math>1,3,5</math> with medians <math>(172,173),(57,58),(34,35)</math> respectively.
  
−
Thus the answer is <math>\textbf{(E) }7</math>.
+
Thus the answer is <math>\boxed{\textbf{(E) }7}</math>.
  
 
==Solution 2==
 
==Solution 2==
  
−
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 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 \cdot (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>.
+
We look for sequences 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 substitute our new value of <math>g</math> into <math>\frac {n \cdot (k+g)}{2}</math> to get that the sum is <math>\frac {n \cdot (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 \cdot (2k+n-1)}{2} = 345</math>. This gives a nice equation. We multiply out the 2 to get that <math>n \cdot (2k+n-1)=690</math>. This leaves us with 2 integers that multiply 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>\boxed{\textbf{(E) }7}</math> ways.
  
 
~~jk23541
 
~~jk23541
 +
 +
 +
An alternate way to finish.
 +
 +
Let
 +
<cmath>
 +
\begin{align*}
 +
2k+n-1 &=\frac{690}{k} \\
 +
n &= k \\
 +
\end{align*}
 +
</cmath>
 +
where <math>k</math> is a factor of <math>690.</math>
 +
We find <math>2k = 1+\frac{690}{k}-k</math> so we need <math>\frac{690}{k} - k</math> to be positive and odd. Fortunately, regardless of the parity of <math>k</math> we see that <math>\frac{690}{k} - k</math> is odd. Furthermore, we need <math>\frac{690}{k} >k</math> which eliminates exact half of the factors. Now, since we need more than <math>1</math> integer to sum up we need <math>k \ge 2</math> which eliminates one more case. There were <math>16</math> cases to begin with, so our answer is <math>\frac{16}{2}-1 = \boxed{\textbf{(E) }7}</math> 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 <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
 +
 +
==Solution 3 (Fast And Clean)==
 +
The median of the sequence <math>m</math> is either an integer or a half integer. Let <math>m=\frac{i}{2}, i \in N</math>, then <math>P=i\cdot n=2\cdot 3 \cdot 5 \cdot 23</math>.
 +
 +
On the other hand we have two constraints:
 +
 +
1) <math>m \geq \frac{n+1}{2} \iff i>n</math> because the integers in the sequence are all positive, and <math>n>1</math>;
 +
 +
2) If <math>n</math> is odd then <math>m</math> is an integer, <math>i</math> is even; if <math>n</math> is even then <math>m</math> is a half integer, <math>i</math> is odd. Therefore, <math>n</math> and <math>i</math> have opposite parity.
 +
 +
Now <math>P</math> has <math>16</math> factors and it is not a perfect square. There are <math>8-1=7</math> choices for <math>1 < n < \sqrt{P}</math>. Also since <math>2|P, 4\nmid P</math>, we know <math>n</math> and <math>\frac{P}{n}</math> must have opposite parity. Therefore the answer is <math>\boxed{\textbf{(E) }7}</math>.
 +
 +
~ asops
 +
 +
==Video Solution 1==
 +
https://youtu.be/dwEm_PcmaYg
 +
 +
~savannahsolver
 +
 +
== Video Solution 2==
 +
https://youtu.be/ZhAZ1oPe5Ds?t=950
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=17|num-a=19}}
 
{{AMC10 box|year=2016|ab=B|num-b=17|num-a=19}}
 +
 +
[[Category:Introductory Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:55, 9 June 2023

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

We look for sequences of $n$ consecutive numbers starting at $k$ and ending at $k+n-1$. We can now substitute $g$ with $k+n-1$. Now we substitute 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 multiply 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

Solution 3 (Fast And Clean)

The median of the sequence $m$ is either an integer or a half integer. Let $m=\frac{i}{2}, i \in N$, then $P=i\cdot n=2\cdot 3 \cdot 5 \cdot 23$.

On the other hand we have two constraints:

1) $m \geq \frac{n+1}{2} \iff i>n$ because the integers in the sequence are all positive, and $n>1$;

2) If $n$ is odd then $m$ is an integer, $i$ is even; if $n$ is even then $m$ is a half integer, $i$ is odd. Therefore, $n$ and $i$ have opposite parity.

Now $P$ has $16$ factors and it is not a perfect square. There are $8-1=7$ choices for $1 < n < \sqrt{P}$. Also since $2|P, 4\nmid P$, we know $n$ and $\frac{P}{n}$ must have opposite parity. Therefore the answer is $\boxed{\textbf{(E) }7}$.

~ asops

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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