Difference between revisions of "2009 AIME I Problems/Problem 8"

Revision as of 14:38, 23 March 2009

Problem 8

Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$ . Consider all possible positive differences of pairs of elements of $S$ . Let $N$ be the sum of all of these differences. Find the remainder when $N$ is divided by $1000$ .

Solution

Solution 1

When computing $N$ , the number $2^x$ will be added $x$ times (for terms $2^x-2^0$ , $2^x-2^1$ , ..., $2^x - 2^{x-1}$ ), and subtracted $10-x$ times. Hence $N$ can be computed as $N=10\cdot 2^{10} + 8\cdot 2^9 + 6\cdot 2^8 + \cdots - 8\cdot 2^1 - 10\cdot 2^0$ .

We can now simply evaluate $N\bmod 1000$ . One reasonably simple way: $\begin{align*} N & = 10(2^{10}-1) + 8(2^9 - 2^1) + 6(2^8-2^2) + 4(2^7-2^3) + 2(2^6-2^4) \\ & = 10(1023) + 8(510) + 6(252) + 4(120) + 2(48) \\ & = 10(1000+23) + 8(500+10) + 6(250+2) + 480 + 96 \\ & \equiv (0 + 230) + (0 + 80) + (500 + 12) + 480 + 96 \\ & \equiv \boxed{398} \end{align*}$

Solution 2

In this solution we show a more general approach that can be used even if $10$ were replaced by a larger value.

As in Solution 1, we show that $N = \sum_{x=0}^{10} (2x-10) 2^x$ .

Let $A = \sum_{x=0}^{10} x2^x$ and let $B=\sum_{x=0}^{10} 2^x$ . Then obviously $N=2A - 10B$ .

Computing $B$ is easy, as this is simply a geometric series with sum $2^{11}-1 = 2047$ . Hence $B\bmod 1000 = 47$ .

We can compute $A$ using a trick known as the change of summation order.

Imagine writing down a table that has rows with labels 0 to 10. In row $x$ , write the number $2^x$ into the first $x$ columns. You will get a triangular table. Obviously, the row sums of this table are of the form $x2^x$ , and therefore the sum of all the numbers is precisely $A$ .

Now consider the ten columns in this table. Let's label them 1 to 10. In column $x$ , you have the values $2^x$ to $2^{10}$ , each of them once. And this is just a geometric series with the sum $2^{11}-2^x$ . We can now sum these column sums to get $A$ . Hence we have $A = (2^{11}-2^1) + (2^{11}-2^2) + \cdots + (2^{11}-2^{10})$ . This simplifies to $10\cdot 2^{11} - (2^1 + 2^2 + \cdots + 2^{10}) = 10\cdot 2^{11} - 2^{11} + 2$ .

Hence $A = 10\cdot 2048 - 2048 + 2 \equiv 480 - 48 + 2 = 434 \pmod{1000}$ .

Then $N = 2A - 10B \equiv 2\cdot 434 - 10\cdot 47 = 868 - 470 = \boxed{398}$ .

Solution 3

Consider the unique differences $2^{a + n} - 2^a$ . Simple casework yields a sum of $\sum_{n = 1}^{10}(2^n - 1)(2^{11 - n} - 1) = \sum_{n = 1}^{10}2^{11} + 1 - 2^n - 2^{11 - n} = 10\cdot2^{11} + 10 - 2(2 + 2^2 + \cdots + 2^{10})$ $= 10\cdot2^{11} + 10 - 2^2(2^{10} - 1)\equiv480 + 10 - 4\cdot23\equiv\boxed{398}\pmod{1000}$ . This method generalizes nicely as well.

@@ Line 44: / Line 44: @@
 Then <math>N = 2A - 10B \equiv 2\cdot 434 - 10\cdot 47 = 868 - 470 = \boxed{398}</math>.
+===Solution 3===
+Consider the unique differences <math>2^{a + n} - 2^a</math>. Simple casework yields a sum of <math>\sum_{n = 1}^{10}(2^n - 1)(2^{11 - n} - 1) = \sum_{n = 1}^{10}2^{11} + 1 - 2^n - 2^{11 - n} = 10\cdot2^{11} + 10 - 2(2 + 2^2 + \cdots + 2^{10})</math>
+<math>= 10\cdot2^{11} + 10 - 2^2(2^{10} - 1)\equiv480 + 10 - 4\cdot23\equiv\boxed{398}\pmod{1000}</math>. This method generalizes nicely as well.
 == See also ==
 {{AIME box|year=2009|n=I|num-b=7|num-a=9}}

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search