Difference between revisions of "2008 AIME II Problems/Problem 1"
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== Solution == | == Solution == | ||
| + | Rewriting this sequence with more terms, we have | ||
| + | <center><cmath>\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*}</cmath></center> | ||
| + | Factoring this expression yields | ||
| + | <center><cmath>\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*}</cmath></center> | ||
| + | Next, we get | ||
| + | <center><cmath>\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*}</cmath></center> | ||
| + | Then, | ||
| + | <center><cmath>\begin{align*} | ||
| + | N &= 2(\frac{(100)(101)}{2} \mbox{, and simplifying, we get}\\ | ||
| + | N &= (100)(101) \mbox{, so}\\ | ||
| + | N &= 10100 \mbox{.} | ||
| + | \end{align*}</cmath></center> | ||
| + | Dividing <math>10100</math> by <math>1000</math> yields a remainder of <math>\boxed{100}</math>. | ||
| + | |||
| + | == Solution 2 == | ||
Since we want the remainder when <math>N</math> is divided by <math>1000</math>, we may ignore the <math>100^2</math> term. Then, applying the [[difference of squares]] factorization to consecutive terms, | Since we want the remainder when <math>N</math> is divided by <math>1000</math>, we may ignore the <math>100^2</math> term. Then, applying the [[difference of squares]] factorization to consecutive terms, | ||
<center><cmath>\begin{align*} | <center><cmath>\begin{align*} | ||
Revision as of 00:33, 13 July 2016
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
Factoring this expression yields
Next, we get
Then,
Dividing 
by yields a remainder of 
.
Solution 2
Since we want the remainder when is divided by , we may ignore the 
term. Then, applying the difference of squares factorization to consecutive terms,
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. 
