diff --git a/2021/day06/Cargo.lock b/2021/day06/Cargo.lock
new file mode 100644
index 0000000..0c65426
--- /dev/null
+++ b/2021/day06/Cargo.lock
@@ -0,0 +1,7 @@
+# This file is automatically @generated by Cargo.
+# It is not intended for manual editing.
+version = 3
+
+[[package]]
+name = "day06"
+version = "0.1.0"
diff --git a/2021/day06/Cargo.toml b/2021/day06/Cargo.toml
new file mode 100644
index 0000000..2f276f3
--- /dev/null
+++ b/2021/day06/Cargo.toml
@@ -0,0 +1,8 @@
+[package]
+name = "day06"
+version = "0.1.0"
+edition = "2021"
+
+# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
+
+[dependencies]
diff --git a/2021/day06/README.md b/2021/day06/README.md
new file mode 100644
index 0000000..69b9e5d
--- /dev/null
+++ b/2021/day06/README.md
@@ -0,0 +1,84 @@
+https://adventofcode.com/2021/day/6
+
+## \--- Day 6: Lanternfish ---
+
+The sea floor is getting steeper. Maybe the sleigh keys got carried this way?
+
+A massive school of glowing
+[lanternfish](https://en.wikipedia.org/wiki/Lanternfish) swims past. They must
+spawn quickly to reach such large numbers - maybe _exponentially_ quickly? You
+should model their growth rate to be sure.
+
+Although you know nothing about this specific species of lanternfish, you make
+some guesses about their attributes. Surely, each lanternfish creates a new
+lanternfish once every _7_ days.
+
+However, this process isn't necessarily synchronized between every lanternfish
+- one lanternfish might have 2 days left until it creates another lanternfish,
+while another might have 4. So, you can model each fish as a single number
+that represents _the number of days until it creates a new lanternfish_.
+
+Furthermore, you reason, a _new_ lanternfish would surely need slightly longer
+before it's capable of producing more lanternfish: two more days for its first
+cycle.
+
+So, suppose you have a lanternfish with an internal timer value of `3`:
+
+  * After one day, its internal timer would become `2`.
+  * After another day, its internal timer would become `1`.
+  * After another day, its internal timer would become `0`.
+  * After another day, its internal timer would reset to `6`, and it would create a _new_ lanternfish with an internal timer of `8`.
+  * After another day, the first lanternfish would have an internal timer of `5`, and the second lanternfish would have an internal timer of `7`.
+
+A lanternfish that creates a new fish resets its timer to `6`, _not`7`_
+(because `0` is included as a valid timer value). The new lanternfish starts
+with an internal timer of `8` and does not start counting down until the next
+day.
+
+Realizing what you're trying to do, the submarine automatically produces a
+list of the ages of several hundred nearby lanternfish (your puzzle input).
+For example, suppose you were given the following list:
+
+[code]
+
+    3,4,3,1,2
+[/code]
+
+This list means that the first fish has an internal timer of `3`, the second
+fish has an internal timer of `4`, and so on until the fifth fish, which has
+an internal timer of `2`. Simulating these fish over several days would
+proceed as follows:
+
+[code]
+
+    Initial state: 3,4,3,1,2
+    After  1 day:  2,3,2,0,1
+    After  2 days: 1,2,1,6,0,8
+    After  3 days: 0,1,0,5,6,7,8
+    After  4 days: 6,0,6,4,5,6,7,8,8
+    After  5 days: 5,6,5,3,4,5,6,7,7,8
+    After  6 days: 4,5,4,2,3,4,5,6,6,7
+    After  7 days: 3,4,3,1,2,3,4,5,5,6
+    After  8 days: 2,3,2,0,1,2,3,4,4,5
+    After  9 days: 1,2,1,6,0,1,2,3,3,4,8
+    After 10 days: 0,1,0,5,6,0,1,2,2,3,7,8
+    After 11 days: 6,0,6,4,5,6,0,1,1,2,6,7,8,8,8
+    After 12 days: 5,6,5,3,4,5,6,0,0,1,5,6,7,7,7,8,8
+    After 13 days: 4,5,4,2,3,4,5,6,6,0,4,5,6,6,6,7,7,8,8
+    After 14 days: 3,4,3,1,2,3,4,5,5,6,3,4,5,5,5,6,6,7,7,8
+    After 15 days: 2,3,2,0,1,2,3,4,4,5,2,3,4,4,4,5,5,6,6,7
+    After 16 days: 1,2,1,6,0,1,2,3,3,4,1,2,3,3,3,4,4,5,5,6,8
+    After 17 days: 0,1,0,5,6,0,1,2,2,3,0,1,2,2,2,3,3,4,4,5,7,8
+    After 18 days: 6,0,6,4,5,6,0,1,1,2,6,0,1,1,1,2,2,3,3,4,6,7,8,8,8,8
+    
+[/code]
+
+Each day, a `0` becomes a `6` and adds a new `8` to the end of the list, while
+each other number decreases by 1 if it was present at the start of the day.
+
+In this example, after 18 days, there are a total of `26` fish. After 80 days,
+there would be a total of `_5934_`.
+
+Find a way to simulate lanternfish. _How many lanternfish would there be after
+80 days?_
+
diff --git a/2021/day06/input/example0 b/2021/day06/input/example0
new file mode 100644
index 0000000..55129f1
--- /dev/null
+++ b/2021/day06/input/example0
@@ -0,0 +1 @@
+3,4,3,1,2
diff --git a/2021/day06/input/example1 b/2021/day06/input/example1
new file mode 100644
index 0000000..09e953e
--- /dev/null
+++ b/2021/day06/input/example1
@@ -0,0 +1,19 @@
+Initial state: 3,4,3,1,2
+After  1 day:  2,3,2,0,1
+After  2 days: 1,2,1,6,0,8
+After  3 days: 0,1,0,5,6,7,8
+After  4 days: 6,0,6,4,5,6,7,8,8
+After  5 days: 5,6,5,3,4,5,6,7,7,8
+After  6 days: 4,5,4,2,3,4,5,6,6,7
+After  7 days: 3,4,3,1,2,3,4,5,5,6
+After  8 days: 2,3,2,0,1,2,3,4,4,5
+After  9 days: 1,2,1,6,0,1,2,3,3,4,8
+After 10 days: 0,1,0,5,6,0,1,2,2,3,7,8
+After 11 days: 6,0,6,4,5,6,0,1,1,2,6,7,8,8,8
+After 12 days: 5,6,5,3,4,5,6,0,0,1,5,6,7,7,7,8,8
+After 13 days: 4,5,4,2,3,4,5,6,6,0,4,5,6,6,6,7,7,8,8
+After 14 days: 3,4,3,1,2,3,4,5,5,6,3,4,5,5,5,6,6,7,7,8
+After 15 days: 2,3,2,0,1,2,3,4,4,5,2,3,4,4,4,5,5,6,6,7
+After 16 days: 1,2,1,6,0,1,2,3,3,4,1,2,3,3,3,4,4,5,5,6,8
+After 17 days: 0,1,0,5,6,0,1,2,2,3,0,1,2,2,2,3,3,4,4,5,7,8
+After 18 days: 6,0,6,4,5,6,0,1,1,2,6,0,1,1,1,2,2,3,3,4,6,7,8,8,8,8
diff --git a/2021/day06/input/input b/2021/day06/input/input
new file mode 100644
index 0000000..1e7a8df
--- /dev/null
+++ b/2021/day06/input/input
@@ -0,0 +1 @@
+5,1,1,4,1,1,4,1,1,1,1,1,1,1,1,1,1,1,4,2,1,1,1,3,5,1,1,1,5,4,1,1,1,2,2,1,1,1,2,1,1,1,2,5,2,1,2,2,3,1,1,1,1,1,1,1,1,5,1,1,4,1,1,1,5,4,1,1,3,3,2,1,1,1,5,1,1,4,1,1,5,1,1,5,1,2,3,1,5,1,3,2,1,3,1,1,4,1,1,1,1,2,1,2,1,1,2,1,1,1,4,4,1,5,1,1,3,5,1,1,5,1,4,1,1,1,1,1,1,1,1,1,2,2,3,1,1,1,1,1,2,1,1,1,1,1,1,2,1,1,1,5,1,1,1,1,4,1,1,1,1,4,1,1,1,1,3,1,2,1,2,1,3,1,3,4,1,1,1,1,1,1,1,5,1,1,1,1,1,1,1,1,4,1,1,2,2,1,2,4,1,1,3,1,1,1,5,1,3,1,1,1,5,5,1,1,1,1,2,3,4,1,1,1,1,1,1,1,1,1,1,1,1,5,1,4,3,1,1,1,2,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,3,3,1,2,2,1,4,1,5,1,5,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,5,1,1,1,4,3,1,1,4
diff --git a/2021/day06/src/main.rs b/2021/day06/src/main.rs
new file mode 100644
index 0000000..ed253b3
--- /dev/null
+++ b/2021/day06/src/main.rs
@@ -0,0 +1,66 @@
+use std::collections::VecDeque;
+use std::io::stdin;
+
+fn main() {
+    println!("Hello, world!");
+
+    let args: Vec<String> = std::env::args().collect();
+
+    if args.len() > 1 && args[1] == "part1" {
+        part1();
+    } else {
+        part2();
+    }
+}
+
+fn part1() {
+    println!("Part 1:");
+
+    let mut input = String::new();
+    stdin().read_line(&mut input).ok();
+    dbg!(&input);
+
+    let count_fishes = count_fishes_after_days(&input, 80);
+    dbg!(count_fishes);
+}
+
+fn part2() {
+    println!("Part 2:");
+
+    let mut input = String::new();
+    stdin().read_line(&mut input).ok();
+    dbg!(&input);
+
+    let count_fishes = count_fishes_after_days(&input, 256);
+    dbg!(count_fishes);
+}
+
+fn count_fishes_after_days(input: &str, after_days: i32) -> i64 {
+    let fishes: Vec<usize> = input
+        .split(',')
+        .map(|s| s.trim())
+        .map(|num| num.parse().unwrap())
+        .collect();
+
+    println!("Fishes: {:?}", fishes);
+
+    let mut buf = VecDeque::from([0; 9]);
+
+    for fish in fishes {
+        buf[fish] += 1
+    }
+    println!("Initial state: {:?}", buf);
+
+    for day in 1..after_days + 1 {
+        buf.rotate_left(1);
+        buf[6] += buf[8];
+        let count_fishes: i64 = buf.iter().sum();
+        println!(
+            "After {:>2} days: {:?}, {:>3} fishes",
+            day, buf, count_fishes
+        );
+    }
+
+    let count_fishes: i64 = buf.iter().sum();
+    return count_fishes;
+}