2011 AIME I Problems/Problem 11

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

2011 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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