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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 26"
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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>.
 
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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== See also ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 25|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 27|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 10:45, 23 July 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

We only need to worry about the decimal expansion of $\sqrt{n^{2}+1}$ since the $-1$ will do nothing to the decimal digits. 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 $2$.

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