Difference between revisions of "2016 AMC 10B Problems/Problem 18"
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==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> 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 |
==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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:38, 28 June 2020
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:
.
So, since it is a sequence of consecutive numbers starting at
and ending at
. We can now substitute
with
. Now we substiute 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 multiplies 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
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
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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