2008 AIME II Problems/Problem 1

Problem

Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$ , where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$ .

Solution 1

Rewriting this sequence with more terms, we have

$\begin{align*} N &= 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + 95^2 - 94^2 - 93^2 + 92^2 + 91^2 + \ldots - 10^2 - 9^2 + 8^2 + 7^2 - 6^2 - 5^2 + 4^2 + 3^2 - 2^2 - 1^2 \mbox{, and reordering, we get}\\ N &= (100^2 - 98^2) + (99^2 - 97^2) + (96^2 - 94^2) + (95^2 - 93^2) + (92^2 - 90^2) + \ldots + (8^2 - 6^2) + (7^2 - 5^2) +(4^2 - 2^2) + (3^2 - 1^2) \mbox{.} \end{align*}$

Factoring this expression yields

$\begin{align*} N &= (100 - 98)(100 + 98) + (99 - 97)(99 + 97) + (96 - 94)(96 + 94) + (95 - 93)(95 + 93) + (92 - 90)(90 + 92) + \ldots + (8 - 6)(8 + 6) + (7 - 5)(7 + 5) + (4 - 2)(4 + 2) + (3 - 1)(3 + 1) \mbox{, leading to}\\ N &= 2(100 + 98) + 2(99 + 97) + 2(96 + 94) + 2(95 + 93) + 2(92 + 90) + \ldots + 2(8 + 6) + 2(7 + 5) + 2(4 + 2) + 2(3 + 1) \mbox{.} \end{align*}$

Next, we get

$\begin{align*} N &= 2(100 + 98 + 99 + 97 + 96 + 94 + 95 + 93 + 92 + 90 + \ldots + 8 + 6 + 7 + 5 + 4 + 2 + 3 + 1 \mbox{, and rearranging terms yields}\\ N &= 2(100 + 99 + 98 + 97 + 96 + \ldots + 5 + 4 + 3 + 2 + 1) \mbox{.} \end{align*}$

Then,

$\begin{align*} N &= 2\left(\frac{(100)(101)}{2}\right) \mbox{, and simplifying, we get}\\ N &= (100)(101) \mbox{, so}\\ N &= 10100 \mbox{.} \end{align*}$

Dividing $10100$ by $1000$ yields a remainder of $\boxed{100}$ .

Solution 2

Since we want the remainder when $N$ is divided by $1000$ , we may ignore the $100^2$ term. Then, applying the difference of squares factorization to consecutive terms,

$\begin{align*} N &= (99-98)(99+98) - (97-96)(97+96) + (95-94)(95 + 94) + \cdots + (3-2)(3+2) - 1 \\ &= \underbrace{197 - 193}_4 + \underbrace{189 - 185}_4 + \cdots + \underbrace{5 - 1}_4 \\ &= 4 \cdot \left(\frac{197-5}{8}+1\right) = \boxed{100} \end{align*}$

Solution 3

By observation, we realize that the sequence $(a+3)^2 + (a+2)^2 - (a+1)^2 - a^2$ alternates every 4 terms. Simplifying, we get $(a+3)^2 + (a+2)^2 - (a+1)^2 - a^2 = 8a + 12$ , turning $N$ into a arithmetic sequence with 25 terms, them being $1, 5, 9, \dots ,97$ , as the series $8a + 12$ alternates every 4 terms.

Applying the sum of arithmetic sequence formula, we get

$\begin{align*} N &= \frac{25\cdot((8\cdot1 + 12) + (8\cdot97 + 12))}{2} \\ &= \frac{25\cdot(20 + 788)}{2} = 10100 \end{align*}$

So the answer would be $\frac{10100}{1000} = \boxed{100}$ .

- erdaifuu

Solution 4

We can remove the $100^2$ since $100^2 \equiv 0 \pmod{1000}$ and use difference of squares to factor out the rest. This gives $(1)(99+98) + (-1)(96+97) + ... +(1)(3+2) +(-1)(1+0)$ Writing this another way, we get $(99(2) - 1) - (97(2)- 1) + (95(2) - 1) - ... +(3(2)-1) - (1(2) -1)$ We know that the last one is negative because all the numbers before multiplying that are in the form $4a - 1$ (eg. $99 = 25(4) - 1$ ) are positive.

Let $x = 99$ . This makes the expression $(2x-1) - (2(x-2) - 1) + (2(x-4) - 1) - ... + (2(x-96) - 1) - (2(x-98) - 1)$ This simplifies to $(2x-1) - (2x-5) + (2x-9) - ... +(2x - 193) - (2x-197)$ Since the first one is positive and the last one is negative, that means there are an even number of terms and using the associative property and the distributive property, all of the $2x$ terms cancel out. A consequence of this is that all of the positive integers turn negative and all the negative ones turn positive(eg. $-(2x-5) = -2x + 5$ ).

We are left with the sequence $-1+5-9+...-193+197$ We can notice the property that the number of terms in the sequence to a positive number n is equal to $(n+3)/4$ , as well as the fact that every pair sums up to 4. Therefore the total number of terms is $(197+3) / 4 = 50$ . Therefore, there are 25 pairs each summing up to 4, leaving us with $25(4) = \boxed{100}$ .

~idk12345678

2008 AIME II (Problems • Answer Key • Resources)
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