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

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== Problem ==
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==Problem==
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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>.
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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==
 
== Solution 1==
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&\equiv 0 \pmod{125}.\end{align*}</cmath>
 
&\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>
 
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>
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== Solution 3 ==
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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
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<cmath>2^a \equiv 2^b \mod 1000 \rightarrow 2^a - 2^b \equiv 0 \mod 1000 </cmath>
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We can factor out a <math>2^b</math> to get
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<cmath>2^b\left(2^{a-b} - 1\right) \equiv 0 \mod 1000</cmath>
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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
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<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
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<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>
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==Solution 4 (Not rigorous)==
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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.
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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>.
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(Everything up to this point has been perfectly rigorous; now, we will destroy this.)
  
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== Solution 4 ==
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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.)
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We know <math>1, 2,</math> and <math>4</math> are in <math>R</math>. Any other element in <math>R</math> must be a multiple of <math>8</math>. All multiples of <math>8</math> under <math>1000</math> sum up to a multiple of <math>1000</math>. So we can ignore them. We need to remove all multiples of <math>5</math>, or <math>40</math> in what we counted because all elements of <math>R</math> can only be divisible by <math>2</math>. But, their sum is also a multiple of <math>1000</math>. Likewise, the sum of all multiples of <math>8k</math> for some <math>k</math> will be a multiple of <math>1000</math>. Thus, our answer is <math>1+2+4=\boxed{007}</math>.
 
  
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Solution by TheUltimate123
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~ihatemath123
  
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== Solution 5 ==
 
  
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Recognize that as you cycle through progressively higher powers of two, you will have to begin repeating in a pattern at some point, since there are only a finite number of possible 3-digit endings and clearly there is no way for a small sub-pattern to form. The previous statement is true by induction since if a pattern began occurring, then it would need to occur for all values before that as well which is clearly untrue unless it encompasses all possible powers of two. Thus we can start thinking about a final value for it to start repeating. It can't be <math>001</math> or <math>501</math>, and it can't be <math>002</math> or <math>502</math> either because the last two digits of any power of 2 greater than <math>2^1</math> are divisible by 4. However, it can be <math>004</math> or <math>504</math>. From the fact that <math>1+2+4+8 \dots 2^n=2^{n+1}-1</math>, we can safely assume that the sum of all possible endings mod 1000 will be <math>2 \times4 -1=\boxed{007}</math>.
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==Solution 5 (Very bad idea)==
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We can simply list out the last <math>3</math> digits of all the powers of <math>2</math> until we find a pattern.
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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
 +
 
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<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>
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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>.
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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>
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~EvanZ
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==Solution 6 (not rigorous) ==
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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>.
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~Vedoral
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==Video Solution ==
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[https://youtu.be/paoxlv_6kI4?si=zQEiMfHQDSGuMtGN 2011 AIME I #11]
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[https://mathproblemsolvingskills.wordpress.com/ MathProblemSolvingSkills.com]
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2011|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2011|n=I|num-b=10|num-a=12}}
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 +
[[Category: Intermediate Number Theory Problems]]
 
{{MAA Notice}}
 
{{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

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