Difference between revisions of "2021 AMC 12A Problems/Problem 16"
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We can look at answer choice <math>C</math>, which is <math>142</math> first. That means that the number of numbers from <math>1</math> to <math>142</math> is roughly the number of numbers from <math>143</math> to <math>200</math>. | We can look at answer choice <math>C</math>, which is <math>142</math> first. That means that the number of numbers from <math>1</math> to <math>142</math> is roughly the number of numbers from <math>143</math> to <math>200</math>. | ||
− | The number of numbers from <math>1</math> to <math>142</math> is <math>\frac{142 | + | The number of numbers from <math>1</math> to <math>142</math> is <math>\frac{142(142+1)}{2}</math> which is approximately <math>10000.</math> The number of numbers from <math>143</math> to <math>200</math> is <math>\frac{200(200+1)}{2}-\frac{142(142+1)}{2}</math> which is approximately <math>10000</math> as well. Therefore, we can be relatively sure the answer choice is <math>\boxed{(C) \text{ } 142}.</math> |
-PureSwag | -PureSwag | ||
Revision as of 13:34, 14 February 2021
- The following problem is from both the 2021 AMC 10A #16 and 2021 AMC 12A #16, so both problems redirect to this page.
Contents
Problem
In the following list of numbers, the integer appears
times in the list for
.
What is the median of the numbers in this list?
Solution 1
There are numbers in total. Let the median be
. We want to find the median
such that
or
Note that
. Plugging this value in as
gives
, so
is the
nd and
rd numbers, and hence, our desired answer.
.
Note that we can derive through the formula
where
is a perfect square less than or equal to
. We set
to
, so
, and
. We then have
. ~approximation by ciceronii
Solution 2
The th number of this sequence is
via the quadratic formula. We can see that if we halve
we end up getting
. This is approximately the number divided by
.
and since
looks like the only number close to it, it is answer
~Lopkiloinm
Solution 3 (answer choices)
We can look at answer choice , which is
first. That means that the number of numbers from
to
is roughly the number of numbers from
to
.
The number of numbers from to
is
which is approximately
The number of numbers from
to
is
which is approximately
as well. Therefore, we can be relatively sure the answer choice is
-PureSwag
Video Solution by Hawk Math
https://www.youtube.com/watch?v=AjQARBvdZ20
Video Solution by OmegaLearn (Using Algebra)
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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.