Difference between revisions of "2016 AMC 10B Problems/Problem 18"
(→Solution 2) |
(→Solution 3 (Fast And Clean)) |
||
(6 intermediate revisions by 6 users not shown) | |||
Line 20: | Line 20: | ||
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>. | 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>. | ||
− | + | 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 \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 | + | 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 | ||
Line 33: | Line 33: | ||
\begin{align*} | \begin{align*} | ||
2k+n-1 &=\frac{690}{k} \\ | 2k+n-1 &=\frac{690}{k} \\ | ||
− | n &= | + | n &= k \\ |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
Line 41: | Line 41: | ||
==Solution 2.1== | ==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 | 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== |
Latest revision as of 16:55, 9 June 2023
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:
.
We look for sequences of consecutive numbers starting at
and ending at
. We can now substitute
with
. Now we substitute 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 multiply 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.
~~jk23541
An alternate way to finish.
Let
where
is a factor of
We find
so we need
to be positive and odd. Fortunately, regardless of the parity of
we see that
is odd. Furthermore, we need
which eliminates exact half of the factors. Now, since we need more than
integer to sum up we need
which eliminates one more case. There were
cases to begin with, so our answer is
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 , and that is because
, which means
and
must be the middle 2 numbers; however, a sequence of length
with middle numbers
and
that consists only of integers would go into the negatives, so any number from 30 onwards wouldn't work, and since
is a trivial, non-counted solution, we get
-ColtsFan10
Solution 3 (Fast And Clean)
The median of the sequence is either an integer or a half integer. Let
, then
.
On the other hand we have two constraints:
1) because the integers in the sequence are all positive, and
;
2) If is odd then
is an integer,
is even; if
is even then
is a half integer,
is odd. Therefore,
and
have opposite parity.
Now has
factors and it is not a perfect square. There are
choices for
. Also since
, we know
and
must have opposite parity. Therefore the answer is
.
~ asops
Video Solution 1
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.