1954 AHSME Problems/Problem 44

Problem 44

A man born in the first half of the nineteenth century was $x$ years old in the year $x^2$. He was born in:

$\textbf{(A)}\ 1849 \qquad \textbf{(B)}\ 1825 \qquad \textbf{(C)}\ 1812 \qquad \textbf{(D)}\ 1836 \qquad \textbf{(E)}\ 1806$

Solution

If a man born in the 19th century was $x$ years of in the year $x^2$, it implies that the year the man was born was $x^2-x$. So, if the man was born in the first half of the 19th century, it means that $x^2-x < 1850$. Noticing that $40^2 - 40 = 1560$ and $50^2-50 = 2450$, we see that $40 < x < 50$. We can guess values until we hit a solution. $43^2-43 = 1806$, so we see that the man had to have been $43$ years old in the year $1849=43^2$, so the answer is $\boxed{\textbf{(E)}}$.