Difference between revisions of "2020 AMC 10A Problems/Problem 11"
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== Solution 4 == | == Solution 4 == | ||
− | We note that <math>44^2 = 1936</math>, which is the first square less than <math>2020</math>, which means that there are 44 addition terms before <math>2020</math>. This makes <math>2020</math> the 2064th term. To find the median, we need the 2020th and 2021th term. We note that every term before <math>2020</math> is one less than the previous term (that is, we subtract 1 to get the previous term). If <math>2020</math> is the 2064th term, than <math>2020 - | + | We note that <math>44^2 = 1936</math>, which is the first square less than <math>2020</math>, which means that there are 44 addition terms before <math>2020</math>. This makes <math>2020</math> the 2064th term. To find the median, we need the 2020th and 2021th term. We note that every term before <math>2020</math> is one less than the previous term (that is, we subtract 1 to get the previous term). If <math>2020</math> is the 2064th term, than <math>2020 - 44</math> is the (2064 - 44)th term. So, the 2020th term is <math>1976</math>. The next term (term 2021) is <math>1977</math>, and the average of these two terms is the median, or <math>\boxed{\textbf{(C) } 1976.5}</math>. |
− | ~ primegn | + | ~ primegn |
==Video Solution== | ==Video Solution== |
Revision as of 00:19, 24 August 2020
- The following problem is from both the 2020 AMC 12A #8 and 2020 AMC 10A #11, so both problems redirect to this page.
Contents
[hide]Problem 11
What is the median of the following list of numbers
Solution 1
We can see that is less than 2020. Therefore, there are
of the
numbers after
. Also, there are
numbers that are under and equal to
. Since
is equal to
, it, with the other squares, will shift our median's placement up
. We can find that the median of the whole set is
, and
gives us
. Our answer is
.
~aryam
Solution 2
As we are trying to find the median of a -term set, we must find the average of the
th and
st terms.
Since is slightly greater than
, we know that the
perfect squares
through
are less than
, and the rest are greater. Thus, from the number
to the number
, there are
terms. Since
is
less than
and
less than
, we will only need to consider the perfect square terms going down from the
th term,
, after going down
terms. Since the
th and
st terms are only
and
terms away from the
th term, we can simply subtract
from
and
from
to get the two terms, which are
and
. Averaging the two, we get
~emerald_block
Solution 3
We want to know the th term and the
st term to get the median.
We know that
So numbers are in between
to
.
So the sum of and
will result in
, which means that
is the
th number.
Also, notice that , which is larger than
.
Then the th term will be
, and similarly the
th term will be
.
Solving for the median of the two numbers, we get
~toastybaker
Solution 4
We note that , which is the first square less than
, which means that there are 44 addition terms before
. This makes
the 2064th term. To find the median, we need the 2020th and 2021th term. We note that every term before
is one less than the previous term (that is, we subtract 1 to get the previous term). If
is the 2064th term, than
is the (2064 - 44)th term. So, the 2020th term is
. The next term (term 2021) is
, and the average of these two terms is the median, or
.
~ primegn
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.