Algebra

Algebra Level 2

Mathematician August De Morgan spent his entire life in the 1800s. In the last year, he says, "I used to be x x years old x 2 x ^ 2 ." In what year was he born?


The answer is 1806.

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2 solutions

Chew-Seong Cheong
Jun 23, 2017

Since De Morgan spent his entire life in the 1800s and was x x years old in the year x 2 x^2 , then:

1800 x 1899 where and denote the ceiling and floor functions respectively. 43 x 43 x = 43 x 2 = 1849 \begin{aligned} \left \lceil \sqrt{1800} \right \rceil & \le x \le \left \lfloor \sqrt{1899} \right \rfloor & \small \color{#3D99F6} \text{where }\lceil \cdot \rceil \text{ and } \lfloor \cdot \rfloor \text{ denote the ceiling and floor functions respectively.} \\ \implies 43 & \le x \le 43 \\ \implies x & = 43 \\ x^2 & = 1849 \end{aligned}

And he was born in the year 1849 43 = 1806 1849-43 = \boxed{1806} .

References:

Maximos Stratis
Jun 23, 2017

He was born in x 2 x x^2 - x year.
We want 1800 < x 2 x < 1900 1800<x^2-x<1900
The only positive integer solution of this inequality is 43 43 .
Therefore he was born in:
4 3 2 43 = 1806 43^2-43=1806



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