University of South Carolina High School Math Contest/1993 Exam/Problem 26


Let $n=1667$. Then the first nonzero digit in the decimal expansion of $\sqrt{n^2 + 1} - n$ is

$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }3 \qquad \mathrm{(D) \ }4 \qquad \mathrm{(E) \ }5$


The given expression is between 0 and 1, so we only need to worry about the decimal expansion of $\sqrt{n^{2}+1}$ and ignore the integer part of the result. We have $(1667^{2}+1)^{1/2}$. Using the extended Binomial Theorem, we have that $\displaystyle {1/2\choose 0} (1667) + {1/2\choose 1} \left(\frac 1{1667}\right) + \cdots$. Now we only have to look at the second term since all the following terms will be too small to affect the first nonzero digit of the decimal expansion. We see that $\frac{1}{2 \cdot 1667}=.00029\ldots$. The answer is $2$.

Alternatively, if you don't know the extended binomial theorem, we can say $\sqrt{n^2 + 1} - n = \epsilon$. Then $\sqrt{n^2 + 1} = n + \epsilon$ so $n^2 + 1 = n^2 + 2n\epsilon + \epsilon^2$, so $\epsilon^2 + 2n\epsilon = 1$. Because $n$ is large, $\epsilon$ is very small, so if we write $\epsilon = \frac{1}{2n} - \frac{\epsilon^2}{2n}$, we may disregard the second term. The result follows as in the previous solution.