Difference between revisions of "1986 AIME Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | 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 = \boxed{033}</math> | + | 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 our answer is <math>\frac{62(63)}{2} + x = 1986 \Longrightarrow x = \boxed{033}</math>. |
== See also == | == See also == | ||
Revision as of 21:01, 10 April 2012
Problem
The pages of a book are numbered 
through 
. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of 
. What was the number of the page that was added twice?
Solution
Denote the page number as 
, with 
. The sum formula shows that 
. Since cannot be very large, disregard it for now and solve 
. The positive root for 
. Quickly testing, we find that 
is too large, but if we plug in 
we find that our answer is 
.
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 | ||
