De Morgan
Algebra
Level
2
In the 1800s, on his birthday, Augustus De Morgan once noted that the square of the age he turned was the year. In which year he was born?
1802
1803
1805
1806
1804
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Note that 1 8 0 0 ≈ 4 2 . 4 and 1 9 0 0 ≈ 4 3 . 6 , so the only year that is a perfect square in the 1800s was 4 3 2 = 1 8 4 9 .
Thus, De Morgan turned 43 in 1849, so he was born in 1 8 4 9 − 4 3 = 1 8 0 6 .
Remark: In fact, De Morgan was born on June 27, 1806.