Sum of remainders of $2^n$

Question

Hints Only for 2011 AIME I Problem 11

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 all elements in $R$. Find the remainder when $S$ is divided by $1000$.

$2^1 \equiv 2, 2^2 \equiv 4, 2^3 \equiv 8, 2^4 \equiv 16, 2^5 \equiv 32, 2^6 \equiv 64, 2^7 \equiv 128, 2^8 \equiv 256, 2^9 \equiv 512$.

I tried many values of $2^x$ to notice a pattern.

$2^{10} \equiv 24 \pmod{1000}$ and $2^{11} = 48 \pmod{1000}$ and this pattern continue until some point, but then stops.

I am not sure if $2^k \pmod{1000}$ ever cycles back.

But since computing $2^k \pmod{1000}$ is pretty quick I went ahead and found:

$2^{12} \equiv 96, 2^{13} \equiv 192, 2^{14} \equiv 184, 2^{15} \equiv 368, 2^{16} \equiv 736, 2^{17} \equiv 472, 2^{18} \equiv 944, 2^{19} \equiv 888, 2^{20} \equiv 776, 2^{21} \equiv 552, 2^{22} \equiv 104, 2^{23} \equiv 208, 2^{24} \equiv 416, 2^{25} \equiv 832, 2^{26} \equiv 664$.

And I see absolutely no recurring pattern. But I do realize it has to stop somewhere since it cant be an infinite sum.

like:

$$2^{4k + n} \equiv 2^{n} \pmod{10}$$

Hints Only

• EDIT

using $a^{\phi(x)} \equiv 1 \pmod{x}$ from the hints given (as comments and answers)

$2^{100} \equiv 1 \pmod{125}$

so:

$2^{100k + n} \equiv 2^n \pmod{125}$

for $\pmod{8}$,

$2 \equiv 2, 2^2 \equiv 4, 2^3 \equiv 0, 2^4 \equiv 0.... \pmod{8}$

So I need:

$$\sum_{k=0}^{99} 2^n = 2^{100} - 1$$

$\sum \pmod{1000}$ is what I need to find:

I have the system:

$$2^{100} - 1 \equiv 0 \pmod{125}, 2^{100} - 1 \equiv 7 \pmod{8}$$

Now, I am in trouble, how to solve a congruence system? (Noob to CRT)

Edit: I am attempting to use CRT.

$125r + 8s = 1$ I need to find a $(r, s)$ ordered (integer) pair.

$r = \frac{1 - 8s}{125} \implies 1 - 8s \equiv 0 \pmod{125} \implies 8s \equiv 1 \pmod{125}$

I am not sure how to proceed.

Answer 1

If $2^i\equiv 2^j\pmod{1000}$, then $2^i\equiv 2^j\pmod{125}$ and $2^i\equiv 2^j\pmod{8}$. But the latter holds for all $i,j$ that are at least $3$. So instead focus on the former. It turns out that the order of $2$, mod $125$, is $100$. That is, $2^{100}\equiv 1\pmod{125}$, but not for any smaller positive (integer) power.

Answer 2

Continuing vadim123's answer, we have that $n\geq 3$ implies $2^n\equiv 0\pmod{8}$. The subgroup generated by $2$ in $\mathbb{Z}_{/125\mathbb{Z}}^*$ is the whole $\mathbb{Z}_{/125\mathbb{Z}}^*$, since: $$ \left|\mathbb{Z}_{/125\mathbb{Z}}^*\right|=100,\quad 2^{50}\equiv -1\pmod{125},\quad 2^{20}\equiv 76\pmod{125} $$ so a power of two $\pmod{125}$ is allowed to be anything but a multiple of five. So for any $n\geq 3$, $2^{n}\pmod{1000}$ is a multiple of eight but not a multiple of five. No further restrictions. Then we just have to compute:

$$\left(2^0+2^1+2^2\right)+\left(2^{3}+2^4+\ldots+2^{102}\right)\pmod{1000}$$ since that sum accounts for every possible remainder. The previous line is $\equiv 7\pmod{8}$ and $7\pmod{125}$, hence the answer is simply $\color{red}{7}$.

Answer 3

(This is a complete solution, not just a hint.)

Preamble: The solutions here (and on AOPS), use the fact that the order of 2 mod 125 is 100. If you're not familiar with how to lift orders, then it can be tedious to calculate/verify by checking certain powers (esp on the AIME if you're time stressed). As it turns out, we do not need that. We can simply use the following:

Lemma: If $\gcd(a(a -1) , n) = 1$, then the sum of the complete residues of $ a^k \pmod{n}$ is always $0 \pmod{n}$.

Proof: Let the order of $a$ be $O$. Then,
$$ 1 + a + a^ 2+ \ldots + a^{O-1} \equiv \frac{ a^O - 1 } { a - 1 } \equiv 0 \pmod{n}.$$

Corollary: In particular for $ a =2$ and odd $n$, the sum of the $2^k$ complete residues will be 0 mod $n$.
Hence, $ S = 1 + 2^1 + 2^2 + (2^3 + \ldots + 2^{O+2} ) \equiv 7 + 0 \pmod{125}$.

We can also easily calculate $ S = 1 + 2^1 + 2^2 + 2^3 + \ldots \equiv 7 \pmod{8}$. Thus, $ S \equiv 7 \pmod{1000}$.