Difference between revisions of "2008 AIME II Problems/Problem 1"
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Let <math>N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2</math>, where the additions and subtractions alternate in pairs. Find the remainder when <math>N</math> is divided by <math>1000</math>. | Let <math>N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2</math>, where the additions and subtractions alternate in pairs. Find the remainder when <math>N</math> is divided by <math>1000</math>. | ||
− | == Solution == | + | == Solution 1 == |
Rewriting this sequence with more terms, we have | Rewriting this sequence with more terms, we have | ||
<center><cmath>\begin{align*} | <center><cmath>\begin{align*} | ||
Line 33: | Line 33: | ||
&= 4 \cdot \left(\frac{197-5}{8}+1\right) = \boxed{100} | &= 4 \cdot \left(\frac{197-5}{8}+1\right) = \boxed{100} | ||
\end{align*}</cmath></center> | \end{align*}</cmath></center> | ||
+ | |||
+ | == Solution 3 == | ||
+ | By observation, we realize that the sequence <cmath>(a+3)^2 + (a+2)^2 - (a+1)^2 - a^2</cmath> alternates every 4 terms. Simplifying, we get <cmath>(a+3)^2 + (a+2)^2 - (a+1)^2 - a^2 = 8a + 12</cmath>, turning <math>N</math> into a arithmetic sequence with 25 terms, them being <math>1, 5, 9, \dots ,97</math>, as the series <math>8a + 12</math> alternates every 4 terms. | ||
+ | |||
+ | Applying the sum of arithmetic sequence formula, we get | ||
+ | <center><cmath>\begin{align*} | ||
+ | N &= \frac{25\cdot((8\cdot1 + 12) + (8\cdot97 + 12))}{2} \ | ||
+ | &= \frac{25\cdot(20 + 788)}{2} = 10100 | ||
+ | \end{align*}</cmath></center> | ||
+ | So the answer would be <cmath>\frac{10100}{1000} = \boxed{100}</cmath>. | ||
+ | |||
+ | |||
+ | - erdaifuu | ||
== See also == | == See also == |
Revision as of 02:20, 19 December 2022
Contents
[hide]Problem
Let , where the additions and subtractions alternate in pairs. Find the remainder when is divided by .
Solution 1
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,
Solution 3
By observation, we realize that the sequence alternates every 4 terms. Simplifying, we get , turning into a arithmetic sequence with 25 terms, them being , as the series alternates every 4 terms.
Applying the sum of arithmetic sequence formula, we get
So the answer would be .
- erdaifuu
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 |
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