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Difference between revisions of "2009 AIME I Problems/Problem 8"
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== Problem 8 ==
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== Problem ==
 
Let <math>S = \{2^0,2^1,2^2,\ldots,2^{10}\}</math>. Consider all possible positive differences of pairs of elements of <math>S</math>. Let <math>N</math> be the sum of all of these differences. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 
Let <math>S = \{2^0,2^1,2^2,\ldots,2^{10}\}</math>. Consider all possible positive differences of pairs of elements of <math>S</math>. Let <math>N</math> be the sum of all of these differences. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  
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==Solution==
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== Solution 1 (bash) ==
  
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=== Solution 1 ===
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When computing <math>N</math>, the number <math>2^x</math> will be added <math>x</math> times (for terms <math>2^x-2^0</math>, <math>2^x-2^1</math>, ..., <math>2^x - 2^{x-1}</math>), and subtracted <math>10-x</math> times. Hence <math>N</math> can be computed as <math>N=10\cdot 2^{10} + 8\cdot 2^9 + 6\cdot 2^8 + \cdots - 8\cdot 2^1 - 10\cdot 2^0</math>. Evaluating <math>N \bmod {1000}</math> yields:
  
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When computing <math>N</math>, the number <math>2^x</math> will be added <math>x</math> times (for terms <math>2^x-2^0</math>, <math>2^x-2^1</math>, ..., <math>2^x - 2^{x-1}</math>), and subtracted <math>10-x</math> times. Hence <math>N</math> can be computed as <math>N=10\cdot 2^{10} + 8\cdot 2^9 + 6\cdot 2^8 + \cdots - 8\cdot 2^1 - 10\cdot 2^0</math>.
 
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We can now simply evaluate <math>N\bmod 1000</math>. One reasonably simple way:
 
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
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</cmath>
 
</cmath>
  
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=== Solution 2 ===
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== Solution 2 ==
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In this solution we show a more general approach that can be used even if <math>10</math> were replaced by a larger value.
 
  
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As in Solution 1, we show that <math>N = \sum_{x=0}^{10} (2x-10) 2^x</math>.  
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This solution can be generalized to apply when <math>10</math> is replaced by other positive integers.
  
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Let <math>A = \sum_{x=0}^{10} x2^x</math> and let <math>B=\sum_{x=0}^{10} 2^x</math>. Then obviously <math>N=2A - 10B</math>.
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Extending from Solution 2, we get that the sum <math>N</math> of all possible differences of pairs of elements in <math>S</math> when <math>S = \{2^0,2^1,2^2,\ldots,2^{n}\}</math> is equal to <math>\sum_{x=0}^{n} (2x-n) 2^x</math>. Let <math>A = \sum_{x=0}^{n} x2^x</math>, <math>B=\sum_{x=0}^{n} 2^x</math>. Then <math>N=2A - nB</math>.
  
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Computing <math>B</math> is easy, as this is simply a geometric series with sum <math>2^{11}-1 = 2047</math>. Hence <math>B\bmod 1000 = 47</math>.
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For <math>n = 10</math>, <math>B = 2^{11}-1 = 2047 \equiv 47 \pmod{1000}</math> by the geometric sequence formula.  
  
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We can compute <math>A</math> using a trick known as the change of summation order.  
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<math>2A = \sum_{x=1}^{n+1} (x-1)2^x</math>, so <math>2A - A = A = n2^{n+1} - \sum_{x=1}^{n} 2^x</math>. Hence, for <math>n = 10</math>, <math>A = 10 \cdot 2^{11} - 2^{11} + 2 = 9 \cdot 2^{11} + 2 \equiv 48 \cdot 9 + 2 =</math>
  
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Imagine writing down a table that has rows with labels 0 to 10. In row <math>x</math>, write the number <math>2^x</math> into the first <math>x</math> columns. You will get a triangular table. Obviously, the row sums of this table are of the form <math>x2^x</math>, and therefore the sum of all the numbers is precisely <math>A</math>.
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<math>434 \pmod{1000}</math>, by the geometric sequence formula and the fact that <math>2^{10} = 1024 \equiv 24 \pmod{1000}</math>.
  
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Now consider the ten columns in this table. Let's label them 1 to 10. In column <math>x</math>, you have the values <math>2^x</math> to <math>2^{10}</math>, each of them once. And this is just a geometric series with the sum <math>2^{11}-2^x</math>. We can now sum these column sums to get <math>A</math>.
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Thus, for <math>n = 10</math>, <math>N = 2A - 10B \equiv 2\cdot 434 - 10\cdot 47 = 868 - 470 = \boxed{398}</math>.
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Hence we have <math>A = (2^{11}-2^1) + (2^{11}-2^2) + \cdots + (2^{11}-2^{10})</math>. This simplifies to <math>10\cdot 2^{11} - (2^1 + 2^2 + \cdots + 2^{10}) = 10\cdot 2^{11} - 2^{11} + 2</math>.
 
  
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Hence <math>A = 10\cdot 2048 - 2048 + 2 \equiv 480 - 48 + 2 = 434 \pmod{1000}</math>.
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==Solution 3==
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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>
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<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.
  
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Then <math>N = 2A - 10B \equiv 2\cdot 434 - 10\cdot 47 = 868 - 470 = \boxed{398}</math>.
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==Solution 4 (Extreme bash)==
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Find the positive differences in all <math>55</math> pairs and you will get <math>\boxed{398}</math>.  
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(This is not recommended unless you can't find any other solutions to this problem)
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2009|n=I|num-b=7|num-a=9}}
 
{{AIME box|year=2009|n=I|num-b=7|num-a=9}}
 +
[[Category:Intermediate Number Theory Problems]]
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[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 10:12, 16 August 2024

Problem

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 1 (bash)

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 $N=10\cdot 2^{10} + 8\cdot 2^9 + 6\cdot 2^8 + \cdots - 8\cdot 2^1 - 10\cdot 2^0$. Evaluating $N \bmod {1000}$ yields:

\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

This solution can be generalized to apply when $10$ is replaced by other positive integers.

Extending from Solution 2, we get that the sum $N$ of all possible differences of pairs of elements in $S$ when $S = \{2^0,2^1,2^2,\ldots,2^{n}\}$ is equal to $\sum_{x=0}^{n} (2x-n) 2^x$. Let $A = \sum_{x=0}^{n} x2^x$, $B=\sum_{x=0}^{n} 2^x$. Then $N=2A - nB$.

For $n = 10$, $B = 2^{11}-1 = 2047 \equiv 47 \pmod{1000}$ by the geometric sequence formula.

$2A = \sum_{x=1}^{n+1} (x-1)2^x$, so $2A - A = A = n2^{n+1} - \sum_{x=1}^{n} 2^x$. Hence, for $n = 10$, $A = 10 \cdot 2^{11} - 2^{11} + 2 = 9 \cdot 2^{11} + 2 \equiv 48 \cdot 9 + 2 =$

$434 \pmod{1000}$, by the geometric sequence formula and the fact that $2^{10} = 1024 \equiv 24 \pmod{1000}$.

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

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.

Solution 4 (Extreme bash)

Find the positive differences in all $55$ pairs and you will get $\boxed{398}$. (This is not recommended unless you can't find any other solutions to this problem)

See also

2009 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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All AIME Problems and Solutions

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