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Difference between revisions of "1986 AIME Problems/Problem 6"

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== Problem ==
 
== Problem ==
 
The pages of a book are numbered <math>1_{}^{}</math> through <math>n_{}^{}</math>. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of <math>1986_{}^{}</math>. What was the number of the page that was added twice?  
 
The pages of a book are numbered <math>1_{}^{}</math> through <math>n_{}^{}</math>. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of <math>1986_{}^{}</math>. What was the number of the page that was added twice?  
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== Solution ==
 
== Solution ==
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{{solution}}
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Denote the page number as <math>x</math>, with <math>x < n</math>. The sum formula shows that <math>\frac{n(n + 1)}{2} + x = 1986</math>. Since <math>x</math> cannot be very large, disregard it for now and solve <math>\frac{n(n+1)}{2} = 1986</math>. The positive root for <math>n \approx \sqrt{3972} \approx 63</math>. Quickly testing, we find that <math>63</math> is too large, but if we plug in <math>62</math> we find that <math>\frac{62(63)}{2} + x = 1986 \Longrightarrow x = 33</math>, our solution.
  
 
== See also ==
 
== See also ==
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* [[1986 AIME Problems]]
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{{AIME box|year=1986|num-b=5|num-a=7}}
  
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{{AIME box|year=1986|num-b=5|num-a=7}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 20:17, 23 March 2007

Problem

The pages of a book are numbered $1_{}^{}$ through $n_{}^{}$. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of $1986_{}^{}$. What was the number of the page that was added twice?

Solution

Denote the page number as $x$, with $x < n$. The sum formula shows that $\frac{n(n + 1)}{2} + x = 1986$. Since $x$ cannot be very large, disregard it for now and solve $\frac{n(n+1)}{2} = 1986$. The positive root for $n \approx \sqrt{3972} \approx 63$. Quickly testing, we find that $63$ is too large, but if we plug in $62$ we find that $\frac{62(63)}{2} + x = 1986 \Longrightarrow x = 33$, our solution.

See also

1986 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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