# June 2020 Brain Puzzler Solution

Q: What is the median of the following list of numbers: (There are 4040 numbers)

1, 2, 3…2020, 1^2, 2^2, 3^2…2020^2?

A: 1976.5

In order to put these into order from least to greatest, we have to figure out how many of the square(d) numbers are less than 2020:

1^2 (1) – 10^2 (100) is 10 numbers
11^2 (121) – 20^2 (400) is 10 numbers
21^2 (441) – 30^2 (900) is 10 numbers
31^2 (961) – 40^2 (1600) is 10 numbers
41^2 (1681) – 44^2 (1936) is 4 numbers

There are 4040 numbers altogether so we need to know the 2020th and 2021st numbers and get the mean of those two numbers to find the median.
Inserting the 44 perfect squares above into the first 2020 numbers would make 2020 the 2064th number so we need to subtract 43 and 44 from 2020 to find those two numbers (the 2020th and 2021st).
By the way, the nearest perfect square under 2020 is 1936 which (luckily!) is too far away for us to think about!

2020 – 43 = 1977 and 2020 – 44 = 1976

Therefore the median of the list of numbers above is 1976.5.