Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AIME I Problems/Problem 11"

(Created page with '== Problem == Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by <math>1000<…')
 
m
 
(63 intermediate revisions by 37 users not shown)
Line 1: Line 1:
−
== Problem ==
+
==Problem==
−
Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by <math>1000</math>. Let <math>S</math> be the sum of the elements in <math>R</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>.
+
 
 +
Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by 1000. Let <math>S</math> be the sum of the elements in <math>R</math>. Find the remainder when <math>S</math> is divided by 1000.
 +
 
 +
== Solution 1==
 +
Note that <math>x \equiv y \pmod{1000} \Leftrightarrow x \equiv y \pmod{125}</math> and <math>x \equiv y \pmod{8}</math>. So we must find the first two integers <math>i</math> and <math>j</math> such that <math>2^i \equiv 2^j \pmod{125}</math> and <math>2^i \equiv 2^j \pmod{8}</math> and <math>i \neq j</math>. Note that <math>i</math> and <math>j</math> will be greater than 2 since remainders of <math>1, 2, 4</math> will not be possible after 2 (the numbers following will always be congruent to 0 modulo 8). Note that <math>2^{100}\equiv 1\pmod{125}</math> (see [[Euler's Totient Theorem|Euler's theorem]]) and <math>2^0,2^1,2^2,\ldots,2^{99}</math> are all distinct modulo 125 (proof below). Thus, <math>i = 103</math> and <math>j =3</math> are the first two integers such that <math>2^i \equiv 2^j \pmod{1000}</math>. All that is left is to find <math>S</math> in mod <math>1000</math>. After some computation:
 +
<cmath>
 +
S = 2^0+2^1+2^2+2^3+2^4+...+2^{101}+ 2^{102} = 2^{103}-1 \equiv 8 - 1 \mod 1000 = \boxed{007}.
 +
</cmath>
 +
To show that <math>2^0, 2^1,\ldots, 2^{99}</math> are distinct modulo 125, suppose for the sake of contradiction that they are not. Then, we must have at least one of <math>2^{20}\equiv 1\pmod{125}</math> or <math>2^{50}\equiv 1\pmod{125}</math>. However, writing <math>2^{10}\equiv 25 - 1\pmod{125}</math>, we can easily verify that <math>2^{20}\equiv -49\pmod{125}</math> and <math>2^{50}\equiv -1\pmod{125}</math>, giving us the needed contradiction.
 +
 
 +
== Solution 2 ==
 +
Notice that our sum of remainders looks like <cmath>S = 1 + 2 + 4 + 2^3 + 2^4 + \cdots.</cmath> We want to find the remainder of <math>S</math> upon division by <math>1000.</math> Since <math>1000</math> decomposes into primes as <math>1000 = 2^3 \cdot 5^3</math>, we can check the remainders of <math>S</math> modulo <math>2^3</math> and modulo <math>5^3</math> separately.
 +
 
 +
Checking <math>S</math> modulo <math>2^3</math> is easy, so lets start by computing the remainder of <math>S</math> upon division by <math>5^3.</math> To do this, let's figure out when our sequence finally repeats.
 +
Notice that since the remainder when dividing any term of <math>S</math> (after the third term) by <math>1000</math> will be a multiple of <math>2^3</math>, when this summation finally repeats, the first term to repeat will be ''not'' be <math>1</math> since <math>2^3 \nmid 1.</math> Instead, the first term to repeat will be <math>2^3</math>, and then the sequence will continue once again <math>2^4, 2^5, \cdots.</math>
 +
 
 +
Now, to compute <math>S</math> modulo <math>125</math>, we want to find the least positive integer <math>d</math> such that <math>2^d \equiv 1 \pmod {125}</math> since then <math>d</math> will just be the number of terms of <math>S</math> (after the third term!) before the sequence repeats. In other words, our sequence will be of the form <math>S = 1 + 2 + 4 + \left(2^3 + 2^4 + \cdots + 2^{2 + d}\right)</math> and then we will have <math>2^{d + 3} \equiv 2^3 \pmod {125}</math>, and the sequence will repeat from there. Here, <math>d</math> simply represents the order of <math>2</math> modulo <math>125</math>, denoted by <math>d = \text{ord}_{125}(2).</math> To begin with, we'll use a well-known property of the order to get a bound on <math>d.</math>
 +
 
 +
Since <math>\gcd(2, 125) = 1</math> and <math>\phi(125) = 100</math>, we know by Euler's Theorem that <math>2^{100} \equiv 1 \pmod {125}.</math> However, we do not know that <math>100</math> is the ''least'' <math>d</math> satisfying <math>2^d \equiv 1 \pmod {125}.</math> Nonetheless, it is a well known property of the order that <math>\text{ord}_{125}(2) = d | \phi(125) = 100.</math> Therefore, we can conclude that <math>d</math> must be a positive divisor of <math>100.</math>
 +
 
 +
Now, this still leaves a lot of possibilities, so let's consider a smaller modulus for the moment, say <math>\mod 5.</math> Clearly, we must have that <math>2^d \equiv 1 \pmod 5.</math> Since <math>2^4 \equiv 1 \pmod 5</math> and powers of two will then cycle every four terms, we know that <math>2^d \equiv 1 \pmod 5 \iff 4 | d.</math> Combining this relation with <math>d | 100</math>, it follows that <math>d \in \{4, 20, 100\}.</math>
 +
 
 +
Now, it is trivial to verify that <math>d \ne 4.</math> In addition, we know that <cmath>2^{20} = \left(2^{10}\right)^2 = \left(1024\right)^2 \equiv 24^2 = 576 \not\equiv 1 \pmod {125}.</cmath> Therefore, we conclude that <math>d \ne 20.</math> Hence, we must have <math>d = 100.</math> (Notice that we could have guessed this by Euler's, but we couldn't have been certain without investigating the order more thoroughly).
 +
 
 +
Now, since we have found <math>d = 100</math>, we know that <cmath>S = 1 + 2 + 4 + 2^3 + 2^4 + \cdots + 2^{102}.</cmath> There are two good ways to finish from here:
 +
 
 +
The first way is to use a trick involving powers of <math>2.</math> Notice that <cmath>S = 1 + 2 + 4 + ... + 2^{102} = 2^{103} - 1.</cmath> Certainly <cmath>S = 2^{103} - 1 \equiv -1 \equiv 7 \pmod {8}.</cmath> In addition, since we already computed <math>\text{ord}_{125}(2) = d = 100</math>, we know that <cmath>S = 2^{103} - 1 = 2^{100} \cdot 2^3 - 1 \equiv 2^3 - 1 \equiv 7 \pmod {125}.</cmath> Therefore, since <math>S \equiv 7 \pmod{8}</math> and <math>S \equiv 7 \pmod{125}</math>, we conclude that <math>S \equiv \boxed{007} \pmod {1000}.</math>
 +
 
 +
The second way is not as slick, but works better in a general setting when there aren't any convenient tricks as in Method 1. Let us split the terms of <math>S</math> into groups: <cmath>R = 1 + 2 + 4; \quad T = 2^3 + 2^4 + \cdots + 2^{102}.</cmath> It is easy to see that <math>R</math> is congruent to <math>7</math> modulo <math>1000.</math>
 +
 
 +
Now, for <math>T</math>, notice that there are <math>100</math> terms in the summation, each with a different remainder upon division by <math>125.</math> Since each of these remainders is certainly relatively prime to <math>125</math>, these <math>100</math> remainders correspond to the <math>100</math> positive integers less than <math>125</math> that are relatively prime to <math>125.</math> Therefore,
 +
<cmath>\begin{align*}T &\equiv 1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 + 11 + \cdots + 124 \pmod{125} \\
 +
&= \left(1 + 2 + 3 + \cdots + 125\right) - \left(5 + 10 + 15 + \cdots 125\right) \\
 +
&= \frac{125 \cdot 126}{2} - 5\left(1 + 2 + 3 + \cdots 25\right) \\
 +
&= 125 \cdot 63 - 5\left(\frac{25 \cdot 26}{2}\right) \\
 +
&= 125\left(63 - 13\right) \\
 +
&\equiv 0 \pmod{125}.\end{align*}</cmath>
 +
Then, since <math>T</math> is divisible by <math>125</math> and <math>8</math>, it follows that <math>T</math> is divisible by <math>1000.</math> Therefore, <cmath>S = R + T \equiv R \equiv \boxed{007} \pmod{1000}.</cmath>
 +
== Solution 3 ==
 +
Obviously, <math>2^i</math> will have to repeat at some point, and our goal is just to find when it repeats. Suppose <math>2^a</math> is the first time the powers of 2 repeat mod 1000, and that it is the same as <math>2^b</math> where <math>b < a.</math> We have
 +
<cmath>2^a \equiv 2^b \mod 1000 \rightarrow 2^a - 2^b \equiv 0 \mod 1000 </cmath>
 +
We can factor out a <math>2^b</math> to get
 +
<cmath>2^b\left(2^{a-b} - 1\right) \equiv 0 \mod 1000</cmath>
 +
Now, let's apply CRT. Obviously, if it is 0 mod 1000, this means directly that it is 0 mod 8 and 0 mod 125. Since <math>2^{a-b} - 1</math> has to be odd, this necessarily means <math>b \ge 3</math> for <math>8 \div 2^b.</math> This means that 125 has to divide <math>2^{a-b} - 1.</math> We wish to find the minimum <math>a - b</math> such that this is true, or similarly, we just wish to find <math>\text{ord}_{125}(2).</math> Note that by Euler's Totient Theorem, that <math>2^{100} \equiv 1 \mod 125.</math> After checking, and seeing that <math>2^{50} \equiv -1 \mod 125</math>, this directly means that <math>\text{ord}_{125}(2) = 100</math> or <math>a - b = 100.</math> Since <math>b \ge 3</math>, the smallest <math>(a, b)</math> pair is
 +
<math>(103, 3).</math> What this really means is that the sequence of remainders will start repeating at <math>2^{103}</math> or namely that
 +
<math>\{2^0, 2^1, ..., 2^{102}\}</math> are all distinct residues mod 1000. Adding these yields <cmath>2^{103} - 1 \equiv 8 - 1 \equiv \boxed{7} \mod 1000</cmath>
 +
==Solution 4 (Not rigorous)==
 +
If we write out the remainders when powers of <math>2</math> are divided by <math>1000</math>, we must eventually write a number we have already written down. After this happens, we will fall into a cycle, and thus, nothing new will be written down.
 +
 
 +
The answer extraction of the problem is equivalent to asking for <math>1+2+4 + \dots + 2^n \pmod{1000}</math>, where <math>2^{n+1}</math> is the first number whose remainder when divided by <math>1000</math> is a repeat. This expression is equivalent to <math>2^{n+1}-1 \pmod{1000}</math>. So our answer will just be the first repeated remainder minus <math>1</math>.
 +
 
 +
(Everything up to this point has been perfectly rigorous; now, we will destroy this.)
 +
 
 +
Note that <math>1</math>, <math>2</math> and <math>4</math> cannot be the first repeated remainders, due to the <math>2</math>, <math>4</math> and <math>8</math> divisibility rules, respectively (i.e. <math>2^0</math>, <math>2^1</math> and <math>2^2</math> are the only powers of <math>2</math> that leave a remainder of <math>1</math>, <math>2</math> and <math>4</math>, respectively, when divided by <math>1000</math>.) There is no reason why <math>8</math> could not be the first repeated remainder, so it probably is. Thus, our answer is <math>8-1 = \boxed{7}</math>. (In fact, one can quite easily show that if <math>8</math> reappears at all in the sequence, it must be the first repeated remainder.)
 +
 
 +
~ihatemath123
 +
 
 +
 
 +
==Solution 5 (Very bad idea)==
 +
We can simply list out the last <math>3</math> digits of all the powers of <math>2</math> until we find a pattern.
 +
 
 +
Note: This solution is very tedious and should not be used unless there is enough remaining time and you can't think of any other way
 +
 
 +
<math>001, 002, 004, 008, 016, 032, 064, 128, 256, 512, 024, 048, 096, 192, 384, 768, 536, 072, 144, 288, 576, 152, 304, 608,</math> <math>216, 432, 864, 728, 456, 912, 824, 648, 296, 592, 184, 368, 736, 472, 944, 888, 776, 552, 104, 208, 416, 832, 664, 328,</math> <math>656, 312, 624, 248, 496, 992 \cdots</math>
 +
 
 +
We can see that after <math>496</math> comes <math>992</math> which can be written as <math>-8 \pmod {1000}</math>. That means that all the terms after <math>8</math> will repeat but negated modulo <math>1000</math>.
 +
Note that <math>n \pmod m + -n \pmod m = 0 \pmod m</math>. As we are looking for the sum of all of the possible values mod <math>1000</math>, the answer is just <math>1 + 2 + 4 = \fbox{007}</math>
 +
 
 +
~EvanZ
 +
 
 +
==Solution 6 (not rigorous) ==
 +
Note that the problem essentially asks for the sum of all <math>x</math> in the residue class of <math>1000</math> that come from a power of <math>2</math>. We can split up this residue class into  finding a solution for <math>x</math> in the modular system <cmath>x \equiv 2^n \bmod 8</cmath> <cmath> x \equiv 2^n \bmod 125</cmath> Clearly, if <math>n \geq 3</math>, we get the first equation to be <cmath>x \equiv 0 \bmod 8</cmath> If <math>n \leq 2</math>, we get our residues to be <math>1</math>,<math>2</math>, and <math>4</math> which add to <math>7</math>. Now, we just need to worry about the <math>125</math> mod equation. If we take a look at the powers of <math>2</math> modulo <math>5</math>, we see that we get every residue except for <math>0</math>. If we consider it modulo <math>25</math>, we see that we get every residue except for <math>0,5,10,15,20</math> (ie. all multiples of <math>5</math>). This is similar to hensel's lemma but it isn't and from here we assume the pattern will continue in that the residues will only be exempt of the multiples of <math>5</math> in modulo <math>125</math>. Therefore, we can add all multiples of <math>8</math> from <math>0</math> to <math>1000</math> and subtract the multiples of <math>40</math> to get <math>0 \bmod 1000</math>. Therefore, our final answer is <math>\fbox{7}</math>.
 +
 
 +
~Vedoral
 +
 
 +
==Video Solution ==
 +
 
 +
[https://youtu.be/paoxlv_6kI4?si=zQEiMfHQDSGuMtGN 2011 AIME I #11]
 +
 
 +
[https://mathproblemsolvingskills.wordpress.com/ MathProblemSolvingSkills.com]
 +
 
 +
== See also ==
 +
{{AIME box|year=2011|n=I|num-b=10|num-a=12}}
 +
 
 +
[[Category: Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:38, 28 August 2024

Problem

Let $R$ be the set of all possible remainders when a number of the form $2^n$, $n$ a nonnegative integer, is divided by 1000. Let $S$ be the sum of the elements in $R$. Find the remainder when $S$ is divided by 1000.

Solution 1

Note that $x \equiv y \pmod{1000} \Leftrightarrow x \equiv y \pmod{125}$ and $x \equiv y \pmod{8}$. So we must find the first two integers $i$ and $j$ such that $2^i \equiv 2^j \pmod{125}$ and $2^i \equiv 2^j \pmod{8}$ and $i \neq j$. Note that $i$ and $j$ will be greater than 2 since remainders of $1, 2, 4$ will not be possible after 2 (the numbers following will always be congruent to 0 modulo 8). Note that $2^{100}\equiv 1\pmod{125}$ (see Euler's theorem) and $2^0,2^1,2^2,\ldots,2^{99}$ are all distinct modulo 125 (proof below). Thus, $i = 103$ and $j =3$ are the first two integers such that $2^i \equiv 2^j \pmod{1000}$. All that is left is to find $S$ in mod $1000$. After some computation: \[S = 2^0+2^1+2^2+2^3+2^4+...+2^{101}+ 2^{102} = 2^{103}-1 \equiv 8 - 1 \mod 1000 = \boxed{007}.\] To show that $2^0, 2^1,\ldots, 2^{99}$ are distinct modulo 125, suppose for the sake of contradiction that they are not. Then, we must have at least one of $2^{20}\equiv 1\pmod{125}$ or $2^{50}\equiv 1\pmod{125}$. However, writing $2^{10}\equiv 25 - 1\pmod{125}$, we can easily verify that $2^{20}\equiv -49\pmod{125}$ and $2^{50}\equiv -1\pmod{125}$, giving us the needed contradiction.

Solution 2

Notice that our sum of remainders looks like \[S = 1 + 2 + 4 + 2^3 + 2^4 + \cdots.\] We want to find the remainder of $S$ upon division by $1000.$ Since $1000$ decomposes into primes as $1000 = 2^3 \cdot 5^3$, we can check the remainders of $S$ modulo $2^3$ and modulo $5^3$ separately.

Checking $S$ modulo $2^3$ is easy, so lets start by computing the remainder of $S$ upon division by $5^3.$ To do this, let's figure out when our sequence finally repeats. Notice that since the remainder when dividing any term of $S$ (after the third term) by $1000$ will be a multiple of $2^3$, when this summation finally repeats, the first term to repeat will be not be $1$ since $2^3 \nmid 1.$ Instead, the first term to repeat will be $2^3$, and then the sequence will continue once again $2^4, 2^5, \cdots.$

Now, to compute $S$ modulo $125$, we want to find the least positive integer $d$ such that $2^d \equiv 1 \pmod {125}$ since then $d$ will just be the number of terms of $S$ (after the third term!) before the sequence repeats. In other words, our sequence will be of the form $S = 1 + 2 + 4 + \left(2^3 + 2^4 + \cdots + 2^{2 + d}\right)$ and then we will have $2^{d + 3} \equiv 2^3 \pmod {125}$, and the sequence will repeat from there. Here, $d$ simply represents the order of $2$ modulo $125$, denoted by $d = \text{ord}_{125}(2).$ To begin with, we'll use a well-known property of the order to get a bound on $d.$

Since $\gcd(2, 125) = 1$ and $\phi(125) = 100$, we know by Euler's Theorem that $2^{100} \equiv 1 \pmod {125}.$ However, we do not know that $100$ is the least $d$ satisfying $2^d \equiv 1 \pmod {125}.$ Nonetheless, it is a well known property of the order that $\text{ord}_{125}(2) = d | \phi(125) = 100.$ Therefore, we can conclude that $d$ must be a positive divisor of $100.$

Now, this still leaves a lot of possibilities, so let's consider a smaller modulus for the moment, say $\mod 5.$ Clearly, we must have that $2^d \equiv 1 \pmod 5.$ Since $2^4 \equiv 1 \pmod 5$ and powers of two will then cycle every four terms, we know that $2^d \equiv 1 \pmod 5 \iff 4 | d.$ Combining this relation with $d | 100$, it follows that $d \in \{4, 20, 100\}.$

Now, it is trivial to verify that $d \ne 4.$ In addition, we know that \[2^{20} = \left(2^{10}\right)^2 = \left(1024\right)^2 \equiv 24^2 = 576 \not\equiv 1 \pmod {125}.\] Therefore, we conclude that $d \ne 20.$ Hence, we must have $d = 100.$ (Notice that we could have guessed this by Euler's, but we couldn't have been certain without investigating the order more thoroughly).

Now, since we have found $d = 100$, we know that \[S = 1 + 2 + 4 + 2^3 + 2^4 + \cdots + 2^{102}.\] There are two good ways to finish from here:

The first way is to use a trick involving powers of $2.$ Notice that \[S = 1 + 2 + 4 + ... + 2^{102} = 2^{103} - 1.\] Certainly \[S = 2^{103} - 1 \equiv -1 \equiv 7 \pmod {8}.\] In addition, since we already computed $\text{ord}_{125}(2) = d = 100$, we know that \[S = 2^{103} - 1 = 2^{100} \cdot 2^3 - 1 \equiv 2^3 - 1 \equiv 7 \pmod {125}.\] Therefore, since $S \equiv 7 \pmod{8}$ and $S \equiv 7 \pmod{125}$, we conclude that $S \equiv \boxed{007} \pmod {1000}.$

The second way is not as slick, but works better in a general setting when there aren't any convenient tricks as in Method 1. Let us split the terms of $S$ into groups: \[R = 1 + 2 + 4; \quad T = 2^3 + 2^4 + \cdots + 2^{102}.\] It is easy to see that $R$ is congruent to $7$ modulo $1000.$

Now, for $T$, notice that there are $100$ terms in the summation, each with a different remainder upon division by $125.$ Since each of these remainders is certainly relatively prime to $125$, these $100$ remainders correspond to the $100$ positive integers less than $125$ that are relatively prime to $125.$ Therefore, \begin{align*}T &\equiv 1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 + 11 + \cdots + 124 \pmod{125} \\ &= \left(1 + 2 + 3 + \cdots + 125\right) - \left(5 + 10 + 15 + \cdots 125\right) \\ &= \frac{125 \cdot 126}{2} - 5\left(1 + 2 + 3 + \cdots 25\right) \\ &= 125 \cdot 63 - 5\left(\frac{25 \cdot 26}{2}\right) \\ &= 125\left(63 - 13\right) \\ &\equiv 0 \pmod{125}.\end{align*} Then, since $T$ is divisible by $125$ and $8$, it follows that $T$ is divisible by $1000.$ Therefore, \[S = R + T \equiv R \equiv \boxed{007} \pmod{1000}.\]

Solution 3

Obviously, $2^i$ will have to repeat at some point, and our goal is just to find when it repeats. Suppose $2^a$ is the first time the powers of 2 repeat mod 1000, and that it is the same as $2^b$ where $b < a.$ We have \[2^a \equiv 2^b \mod 1000 \rightarrow 2^a - 2^b \equiv 0 \mod 1000\] We can factor out a $2^b$ to get \[2^b\left(2^{a-b} - 1\right) \equiv 0 \mod 1000\] Now, let's apply CRT. Obviously, if it is 0 mod 1000, this means directly that it is 0 mod 8 and 0 mod 125. Since $2^{a-b} - 1$ has to be odd, this necessarily means $b \ge 3$ for $8 \div 2^b.$ This means that 125 has to divide $2^{a-b} - 1.$ We wish to find the minimum $a - b$ such that this is true, or similarly, we just wish to find $\text{ord}_{125}(2).$ Note that by Euler's Totient Theorem, that $2^{100} \equiv 1 \mod 125.$ After checking, and seeing that $2^{50} \equiv -1 \mod 125$, this directly means that $\text{ord}_{125}(2) = 100$ or $a - b = 100.$ Since $b \ge 3$, the smallest $(a, b)$ pair is $(103, 3).$ What this really means is that the sequence of remainders will start repeating at $2^{103}$ or namely that $\{2^0, 2^1, ..., 2^{102}\}$ are all distinct residues mod 1000. Adding these yields \[2^{103} - 1 \equiv 8 - 1 \equiv \boxed{7} \mod 1000\]

Solution 4 (Not rigorous)

If we write out the remainders when powers of $2$ are divided by $1000$, we must eventually write a number we have already written down. After this happens, we will fall into a cycle, and thus, nothing new will be written down.

The answer extraction of the problem is equivalent to asking for $1+2+4 + \dots + 2^n \pmod{1000}$, where $2^{n+1}$ is the first number whose remainder when divided by $1000$ is a repeat. This expression is equivalent to $2^{n+1}-1 \pmod{1000}$. So our answer will just be the first repeated remainder minus $1$.

(Everything up to this point has been perfectly rigorous; now, we will destroy this.)

Note that $1$, $2$ and $4$ cannot be the first repeated remainders, due to the $2$, $4$ and $8$ divisibility rules, respectively (i.e. $2^0$, $2^1$ and $2^2$ are the only powers of $2$ that leave a remainder of $1$, $2$ and $4$, respectively, when divided by $1000$.) There is no reason why $8$ could not be the first repeated remainder, so it probably is. Thus, our answer is $8-1 = \boxed{7}$. (In fact, one can quite easily show that if $8$ reappears at all in the sequence, it must be the first repeated remainder.)

~ihatemath123


Solution 5 (Very bad idea)

We can simply list out the last $3$ digits of all the powers of $2$ until we find a pattern.

Note: This solution is very tedious and should not be used unless there is enough remaining time and you can't think of any other way

$001, 002, 004, 008, 016, 032, 064, 128, 256, 512, 024, 048, 096, 192, 384, 768, 536, 072, 144, 288, 576, 152, 304, 608,$ $216, 432, 864, 728, 456, 912, 824, 648, 296, 592, 184, 368, 736, 472, 944, 888, 776, 552, 104, 208, 416, 832, 664, 328,$ $656, 312, 624, 248, 496, 992 \cdots$

We can see that after $496$ comes $992$ which can be written as $-8 \pmod {1000}$. That means that all the terms after $8$ will repeat but negated modulo $1000$. Note that $n \pmod m + -n \pmod m = 0 \pmod m$. As we are looking for the sum of all of the possible values mod $1000$, the answer is just $1 + 2 + 4 = \fbox{007}$

~EvanZ

Solution 6 (not rigorous)

Note that the problem essentially asks for the sum of all $x$ in the residue class of $1000$ that come from a power of $2$. We can split up this residue class into finding a solution for $x$ in the modular system \[x \equiv 2^n \bmod 8\] \[x \equiv 2^n \bmod 125\] Clearly, if $n \geq 3$, we get the first equation to be \[x \equiv 0 \bmod 8\] If $n \leq 2$, we get our residues to be $1$,$2$, and $4$ which add to $7$. Now, we just need to worry about the $125$ mod equation. If we take a look at the powers of $2$ modulo $5$, we see that we get every residue except for $0$. If we consider it modulo $25$, we see that we get every residue except for $0,5,10,15,20$ (ie. all multiples of $5$). This is similar to hensel's lemma but it isn't and from here we assume the pattern will continue in that the residues will only be exempt of the multiples of $5$ in modulo $125$. Therefore, we can add all multiples of $8$ from $0$ to $1000$ and subtract the multiples of $40$ to get $0 \bmod 1000$. Therefore, our final answer is $\fbox{7}$.

~Vedoral

Video Solution

2011 AIME I #11

MathProblemSolvingSkills.com

See also

2011 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_11&oldid=225905"