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

Latest revision as of 13:13, 6 September 2020

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

Solution 1

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 our answer is $\frac{62(63)}{2} + x = 1986 \Longrightarrow x = \boxed{033}$ .

Solution 2

Use the same method as above where you represent the sum of integers from $1$ to $n$ expressed as $\frac{n(n + 1)}{2}$ , plus the additional page number $k$ . We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.)

$\frac{n(n + 1)}{2} = 1986$ is the quadratic you must solve to obtain the upper bound of $n$ and $\frac{n(n + 1)}{2} + n = 1986$ is the quadratic you must solve to obtain the lower bound of $n$ .

Solving the two equations gives values that are respectively around $62.5$ and $61.5$ with the quadratic formula, and the only integer between the two is $62$ .

This implies that we can plug in $62$ and come to the same conclusion as the above solution where $x = \boxed{033}$ .

@@ Line 1: / Line 1: @@
 == 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?
-== Solution ==
+==Solution==
+==Solution 1==
+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>.
+==Solution 2==
+Use the same method as above where you represent the sum of integers from <math>1</math> to <math>n</math> expressed as <math>\frac{n(n + 1)}{2}</math>, plus the additional page number <math>k</math>. We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.)
+<math>\frac{n(n + 1)}{2} = 1986</math> is the quadratic you must solve to obtain the upper bound of <math>n</math> and <math>\frac{n(n + 1)}{2} + n = 1986</math>
+is the quadratic you must solve to obtain the lower bound of <math>n</math>.
+Solving the two equations gives values that are respectively around <math>62.5</math> and <math>61.5</math> with the quadratic formula, and the only integer between the two is <math>62</math>.
+This implies that we can plug in <math>62</math> and come to the same conclusion as the above solution where <math>x = \boxed{033}</math>.
 == See also ==
-* [[1984 AIME Problems]]
+{{AIME box|year=1986|num-b=5|num-a=7}}
+* [[AIME Problems and Solutions]]
+* [[American Invitational Mathematics Examination]]
+* [[Mathematics competition resources]]
+[[Category:Intermediate Number Theory Problems]]
+{{MAA Notice}}

1986 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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