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

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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>.
  
==Solution==
+
==Solutions==
We can do this in an organized way.
 
<math> (2^{10}-2^9)+(2^{10}-2^8) ... +(2^{10}-2^0) = (10)2^{10} - 2^9-2^8 ... -2^0</math>
 
<math> (2^9-2^8)+(2^9-2^7) ...+(2^9-2^0) = (9)2^9 - 2^8-2^7 ... -2^0</math>
 
<math>(2^8-2^7)+(2^8-2^6) ...+(2^8-2^0) = (8)2^8-2^7-2^6 ... -2^0</math>
 
  
If we continue doing this, we will have
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===Solution 1===
<math>(10)2^{10}+(9)2^9+(8)2^8 ... +2^1 -2^9-(2)2^8-(3)2^7 ... -(10)2^0</math>
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Find the positive differences in all <math>55</math> pairs and you will get <math>\boxed{398}</math>.
  
Which is
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=== Solution 2 ===
<math>(10)2^{10}+(8)2^9+(6)2^8+(4)2^7 ... +(-10)2^0</math>
 
  
By simplifying this, we will get
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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>.
<math>(16)2^{10}+2^7-(7)2^4-2</math>
 
  
We only care about the last three digits
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We can now simply evaluate <math>N\bmod 1000</math>. One reasonably simple way:
so the answer will be
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<cmath>
<math>384+128-112-2 = 398</math>
+
\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*}
 +
</cmath>
 +
 
 +
=== Solution 3 ===
 +
 
 +
In this solution we show a more general approach that can be used even if <math>10</math> were replaced by a larger value.
 +
 
 +
As in Solution 1, we show that <math>N = \sum_{x=0}^{10} (2x-10) 2^x</math>.
 +
 
 +
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>.
 +
 
 +
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>.
 +
 
 +
We can compute <math>A</math> 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 <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>.
 +
 
 +
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>.
 +
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>.
 +
 
 +
Hence <math>A = 10\cdot 2048 - 2048 + 2 \equiv 480 - 48 + 2 = 434 \pmod{1000}</math>.
 +
 
 +
Then <math>N = 2A - 10B \equiv 2\cdot 434 - 10\cdot 47 = 868 - 470 = \boxed{398}</math>.
 +
 
 +
===Solution 4===
 +
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.
 +
 
 +
==Video Solution==
 +
https://youtu.be/JIVs2eexmVQ
 +
 
 +
== See also ==
 +
{{AIME box|year=2009|n=I|num-b=7|num-a=9}}
 +
[[Category:Intermediate Number Theory Problems]]
 +
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Revision as of 20:11, 19 August 2020

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 $1000$.

Solutions

Solution 1

Find the positive differences in all $55$ pairs and you will get $\boxed{398}$.

Solution 2

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 3

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 4

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.

Video Solution

https://youtu.be/JIVs2eexmVQ

See also

2009 AIME I (ProblemsAnswer KeyResources)
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

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