$2 \cdot 72$ A person was born in the first half of the 19th century, and in the year $x^{2}$, their age was $x$. Therefore, the birth year of this person is
(A) 1849.
(B) 1825.
(C) 1812.
(D) 1836.
(E) 1806.
(5th American High School Mathematics Examination, 1954)
[Solution] Given $1800 < x^2 < 1900$, which means $10 \sqrt{18} < x < 10 \sqrt{19}$, and $10 \sqrt{18} \approx 42$, $10 \sqrt{19} \approx 43$.

Since $x$ is an integer, by checking, when $x = 43$, $x^2 = 1849$, and $1849 - 43 = 1806$. Therefore, the answer is $(E)$.