2011 AIME I Problems/Problem 11

Problem

What is the sum of all positive integers smaller than 1000 that can be written in the form 100^2n , where is an integer (not necessarily positive)?

Solution 1

Note that the invariance of $R \subseteq \mathbb Z_{1000}$ upon multiplication by $2$ implies the invariance of $R \setminus \{1, 2, 4\}$ as $2^n \equiv 0 \mod{8}$ for $n \ge 3.$ Thus, letting the sum of the residues less $1, 2, 4$ be $S',$ we have $2S' \equiv S' \mod{1000} \Rightarrow S' \equiv 0 \mod{1000},$ from which $S = S' + 1 + 2 + 4 \equiv \boxed{007} \mod{1000}$ follows.

Solution 2

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 3

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 4

We know $1, 2,$ and $4$ are in $R$ . Any other element in $R$ must be a multiple of $8$ . All multiples of $8$ under $1000$ sum up to a multiple of $1000$ . So we can ignore them. We need to remove all multiples of $5$ , or $40$ in what we counted because all elements of $R$ can only be divisible by $2$ . But, their sum is also a multiple of $1000$ . Likewise, the sum of all multiples of $8k$ for some $k$ will be a multiple of $1000$ . Thus, our answer is $1+2+4=\boxed{007}$ .

Solution by TheUltimate123

Solution 5

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 $001$ or $501$ , and it can't be $002$ or $502$ either because the last two digits of any power of 2 greater than $2^1$ are divisible by 4. However, it can be $004$ or $504$ . From the fact that $1+2+4+8 \dots 2^n=2^{n+1}-1$ , we can safely assume that the sum of all possible endings mod 1000 will be $2 \times4 -1=\boxed{007}$ .

Solution 6 (assumptions)

We know that units digits repeat every $4$ , and tens digits every $20$ . Therefore, continuing the pattern, hundreds digits should repeat every $100$ . But taking modulo $8$ , we know that $1, 2, 4$ will not be the same modulo $1000$ . Hence, we start at $2^3$ and move up $100$ . That lands us at the first repeat, $2^{103}$ . Just sum up to $2^{102}$ to get a modulus of $\boxed{007}$ .

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

Page

Toolbox

Search