Problem:
What is the median of the following list of 4040 numbers?
1,2,3,…,2020,12,22,32,…,20202
Answer Choices:
A. 1974.5
B. 1975.5
C. 1976.5
D. 1977.5
E. 1978.5
Solution:
We can see that
442=1936 which is less than 2020
Therefore, there are
2020−44=1976 of the 4040 numbers greater than 2020
Also, there are
2020+44=2064 numbers that are less than or equal to 2020
Since there are 44 duplicates/extras, it will shift up our median's placement down 44. Had the list of numbers been 1,2,3,…,4040, the median of the whole set would be
21+4040​=2020.5
Thus, our answer is
2020.5−44=(C) 1976.5​.
OR
As we are trying to find the median of a 4040-term set, we must find the average of the 2020th and 2021st terms Since 452=2025 is slightly greater than 2020, we know that the 44 perfect squares 12 through 442 are less than 2020, and the rest are greater. Thus, from the number 1 to the number 2020, there are 2020+44=2064 terms. Since 442 is 44+45=89 less than 452=2025 and 84 less than 2020, we will only need to consider the perfect square terms going down from the 2064th term, 2020, after going down 84 terms.
Since the 2020th and 2021st terms are only 44 and 43 terms away from the 2064th term, we can simply subtract 44 from 2020 and 43 from 2020 to get the two terms, which are 1976 and 1977.
Averaging the two, we get (C) 1976.5​.
The problems on this page are the property of the MAA's American Mathematics Competitions