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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 26"
 
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== Solution ==
 
== Solution ==
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We only need to worry about the decimal expansion of <math>\sqrt{n^{2}+1}</math> since the <math>-1</math> will do nothing to the decimal digits. We have <math>(1667^{2}+1)^{1/2}</math>. Using the extended binomial theorem, we have that <math> \displaystyle {1/2\choose 0} (1667) + {1/2\choose 1} \left(\frac 1{1667}\right) + \cdots </math>. 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 <math>\frac{1}{2 \cdot 1667}=.00029\ldots</math>. The answer is <math>2</math>.
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The given [[expression]] is between 0 and 1, so we only need to worry about the [[decimal expansion]] of <math>\sqrt{n^{2}+1}</math> and ignore the integer part of the result. We have <math>(1667^{2}+1)^{1/2}</math>. Using the extended [[Binomial Theorem]], we have that <math> \displaystyle {1/2\choose 0} (1667) + {1/2\choose 1} \left(\frac 1{1667}\right) + \cdots </math>. 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 <math>\frac{1}{2 \cdot 1667}=.00029\ldots</math>. The answer is <math>2</math>.
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Alternatively, if you don't know the extended binomial theorem, we can say <math>\sqrt{n^2 + 1} - n = \epsilon</math>.  Then <math>\sqrt{n^2 + 1} = n + \epsilon</math> so <math>n^2 + 1 = n^2 + 2n\epsilon + \epsilon^2</math>, so <math>\epsilon^2 + 2n\epsilon = 1</math>.  Because <math>n</math> is large, <math>\epsilon</math> is very small, so if we write <math>\epsilon = \frac{1}{2n} - \frac{\epsilon^2}{2n}</math>, we may disregard the second term.  The result follows as in the previous solution.
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Latest revision as of 16:34, 18 August 2006

Problem

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

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

Solution

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.


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