Difference between revisions of "2008 AIME II Problems/Problem 1"
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<center><cmath>\begin{align*} | <center><cmath>\begin{align*} | ||
− | N &= 2(\frac{(100)(101)}{2} \mbox{, and simplifying, we get}\\ | + | N &= 2\left(\frac{(100)(101)}{2}\right) \mbox{, and simplifying, we get}\\ |
N &= (100)(101) \mbox{, so}\\ | N &= (100)(101) \mbox{, so}\\ | ||
N &= 10100 \mbox{.} | N &= 10100 \mbox{.} |
Revision as of 22:10, 26 June 2017
Contents
Problem
Let , where the additions and subtractions alternate in pairs. Find the remainder when
is divided by
.
Solution
Rewriting this sequence with more terms, we have
![\begin{align*} N &= 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + 95^2 - 94^2 - 93^2 + 92^2 + 91^2 + \ldots - 10^2 - 9^2 + 8^2 + 7^2 - 6^2 - 5^2 + 4^2 + 3^2 - 2^2 - 1^2 \mbox{, and reordering, we get}\\ N &= (100^2 - 98^2) + (99^2 - 97^2) + (96^2 - 94^2) + (95^2 - 93^2) + (92^2 - 90^2) + \ldots + (8^2 - 6^2) + (7^2 - 5^2) +(4^2 - 2^2) + (3^2 - 1^2) \mbox{.} \end{align*}](http://latex.artofproblemsolving.com/a/7/9/a795bfc297226b5a2b580351d5a15c093eb2213a.png)
Factoring this expression yields
![\begin{align*} N &= (100 - 98)(100 + 98) + (99 - 97)(99 + 97) + (96 - 94)(96 + 94) + (95 - 93)(95 + 93) + (92 - 90)(90 + 92) + \ldots + (8 - 6)(8 + 6) + (7 - 5)(7 + 5) + (4 - 2)(4 + 2) + (3 - 1)(3 + 1) \mbox{, leading to}\\ N &= 2(100 + 98) + 2(99 + 97) + 2(96 + 94) + 2(95 + 93) + 2(92 + 90) + \ldots + 2(8 + 6) + 2(7 + 5) + 2(4 + 2) + 2(3 + 1) \mbox{.} \end{align*}](http://latex.artofproblemsolving.com/0/3/2/0325a66a6a77614ee82a197b2542e32b71487a4d.png)
Next, we get
![\begin{align*} N &= 2(100 + 98 + 99 + 97 + 96 + 94 + 95 + 93 + 92 + 90 + \ldots + 8 + 6 + 7 + 5 + 4 + 2 + 3 + 1 \mbox{, and rearranging terms yields}\\ N &= 2(100 + 99 + 98 + 97 + 96 + \ldots + 5 + 4 + 3 + 2 + 1) \mbox{.} \end{align*}](http://latex.artofproblemsolving.com/5/a/c/5ac9cce92dfbc2447fb71967dd509cdf51a978e2.png)
Then,
![\begin{align*} N &= 2\left(\frac{(100)(101)}{2}\right) \mbox{, and simplifying, we get}\\ N &= (100)(101) \mbox{, so}\\ N &= 10100 \mbox{.} \end{align*}](http://latex.artofproblemsolving.com/f/2/3/f23bce534975df5865cc4154bcc8a877e532fe5f.png)
Dividing by
yields a remainder of
.
Solution
Since we want the remainder when is divided by
, we may ignore the
term. Then, applying the difference of squares factorization to consecutive terms,
![\begin{align*} N &= (99-98)(99+98) - (97-96)(97+96) + (95-94)(95 + 94) + \cdots + (3-2)(3+2) - 1 \\ &= \underbrace{197 - 193}_4 + \underbrace{189 - 185}_4 + \cdots + \underbrace{5 - 1}_4 \\ &= 4 \cdot \left(\frac{197-5}{8}+1\right) = \boxed{100} \end{align*}](http://latex.artofproblemsolving.com/0/b/1/0b1a194ec0e04de5792553302910e506b8737c33.png)
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.