diff --git a/problem-1/problem-1.test.js b/problem-1/problem-1.test.js
index c775a02..4006f53 100644
--- a/problem-1/problem-1.test.js
+++ b/problem-1/problem-1.test.js
@@ -1,4 +1,50 @@
+// 1. 가장 익숙한 방법으로 문제를 해결해 주세요.
+// const solution = (numbers) => {
+//  let answer = 0;
+//
+//  for (let i = 0; i < numbers.length; i += 1) {
+//    answer += numbers[i];
+//  }
+//
+//  return answer;
+// };
+
+// 2. 이번에는 재귀 함수로 문제를 해결해 주세요.
+// const solution = (numbers) => {
+//  // Base Case
+//  if (numbers.length === 0) {
+//    return 0;
+//  }
+//
+//  // Recursive Case
+//  return numbers.pop() + solution(numbers);
+// };
+
+// 3. 꼬리 재귀 함수로 바꿔보세요.
+// -> Maximum call stack size exceeded 발생..
+// const solution = (numbers, acc = 0) => {
+//  if (numbers.length === 0) {
+//    return acc;
+//  }
+//
+//  // return solution(numbers.slice(0, numbers.length - 1), acc + numbers[numbers.length - 1]);
+//  return solution(numbers, acc + numbers.pop());
+// };
+
+// 4. 꼬리 재귀 최적화를 통해서 최적화해 보세요.
 const solution = (numbers) => {
+  // 1) 계속해서 바뀌는 값을 변수로 선언한다.
+  let acc = 0;
+  let index = 0;
+
+  // 3) 반복문을 만든다.
+  while (index < numbers.length) {
+    // 2) 재귀 함수 호출로 값을 변경하는 대신에 변수의 값을 할당해서 변경한다.
+    acc = acc + numbers[index];
+    index += 1;
+  }
+
+  return acc;
 };
 
 test('빈 배열은 0을 반환한다', () => {
diff --git a/problem-2/problem-2.test.js b/problem-2/problem-2.test.js
index 458a6b8..e39822e 100644
--- a/problem-2/problem-2.test.js
+++ b/problem-2/problem-2.test.js
@@ -1,4 +1,67 @@
+// 1. 가장 익숙한 방법으로 문제를 해결해 주세요.
+// const solution = (n) => {
+//  if (n <= 0) {
+//    return 0;
+//  }
+//
+//  if (n === 1) {
+//    return 1;
+//  }
+//
+//  const fibonacci = [0, 1];
+//
+//  for (let i = 2; i <= n; i += 1) {
+//    fibonacci[i] = fibonacci[i - 2] + fibonacci[i - 1];
+//  }
+//
+//  return fibonacci[n];
+// };
+
+// 2. 이번에는 재귀 함수로 문제를 해결해 주세요.
+// const solution = (n) => {
+//  if (n <= 0) {
+//    return 0;
+//  }
+//
+//  if (n === 1) {
+//    return 1;
+//  }
+//
+//  return solution(n - 2) + solution(n - 1);
+// };
+
+// 3. 꼬리 재귀 함수로 바꿔보세요.
+// const solution = (n, prev = 0, curr = 1) => {
+//  if (n <= 0) {
+//    return prev;
+//  }
+//
+//  if (n === 1) {
+//    return curr;
+//  }
+//
+//  return solution(n - 1, curr, prev + curr);
+// };
+
+// 4. 꼬리 재귀 최적화를 통해서 최적화해 보세요.
 const solution = (n) => {
+  let currentNumber = n;
+  let previousValue = 0;
+  let currentValue = 1;
+  let nextValue;
+
+  while (currentNumber > 0) {
+    if (currentNumber === 1) {
+      return currentValue;
+    }
+
+    nextValue = previousValue + currentValue;
+    previousValue = currentValue;
+    currentValue = nextValue;
+    currentNumber -= 1;
+  }
+
+  return previousValue;
 };
 
 test('음수가 주어지면 0을 반환한다', () => {
diff --git a/problem-3/problem-3.test.js b/problem-3/problem-3.test.js
index 7319eb8..3239e1e 100644
--- a/problem-3/problem-3.test.js
+++ b/problem-3/problem-3.test.js
@@ -1,4 +1,30 @@
-const solution = (n) => {
+// const solution = (n) => {
+//  if (n === 0) {
+//    return '0';
+//  }
+//
+//  let number = n;
+//  let binaryString = '';
+//
+//  while (number > 0) {
+//    const remainder = number % 2;
+//    binaryString = remainder + binaryString;
+//    number = Math.floor(number / 2);
+//  }
+//
+//  return binaryString;
+// };
+
+const solution = (n, binaryString = '') => {
+  if (n === 0) {
+    return '0';
+  }
+
+  if (n === 1) {
+    return `1${binaryString}`;
+  }
+
+  return solution(Math.floor(n / 2), (n % 2).toString() + binaryString);
 };
 
 test('이진수 문자열을 반환한다', () => {
diff --git a/problem-4/problem-4.test.js b/problem-4/problem-4.test.js
index 19f5cb0..7cfeb63 100644
--- a/problem-4/problem-4.test.js
+++ b/problem-4/problem-4.test.js
@@ -1,4 +1,67 @@
-const solution = () => {
+// const solution = (binaryString) => {
+//  let decimalNumber = 0;
+//  for (let bitIndex = 0; bitIndex < binaryString.length; bitIndex += 1) {
+//    const currentBit = parseInt(binaryString[bitIndex], 10);
+//    const bitWeight = 2 ** (binaryString.length - 1 - bitIndex);
+//
+//    const currentBitValue = currentBit * bitWeight;
+//    decimalNumber += currentBitValue;
+//  }
+//
+//  return decimalNumber;
+// };
+
+// const solution = (
+//  binaryString,
+//  currentIndex = 0,
+//  bitWeight = binaryString.length - 1,
+// ) => {
+//  if (binaryString === '0') {
+//    return 0;
+//  }
+//
+//  if (binaryString === '1') {
+//    return 1;
+//  }
+//
+//  if (bitWeight < 0) {
+//    return 0;
+//  }
+//
+//  const currentBit = parseInt(binaryString[currentIndex], 10);
+//  const currentBitValue = (2 ** bitWeight) * currentBit;
+//
+//  return currentBitValue + solution(
+//    binaryString,
+//    currentIndex + 1,
+//    bitWeight - 1,
+//  );
+// };
+
+const solution = (
+  binaryString,
+  currentIndex = 0,
+  bitWeight = binaryString.length - 1,
+  decimalNumber = 0,
+) => {
+  if (binaryString === '0') {
+    return 0;
+  }
+
+  if (binaryString === '1') {
+    return 1;
+  }
+
+  if (bitWeight < 0) {
+    return decimalNumber;
+  }
+
+  return solution(
+    binaryString,
+    currentIndex + 1,
+    bitWeight - 1,
+    decimalNumber + (2 ** bitWeight) * parseInt(binaryString[currentIndex], 10),
+  );
 };
 
 test('10진수 숫자를 반환한다', () => {
diff --git a/problem-5/problem-5.test.js b/problem-5/problem-5.test.js
index 20a8fab..e59e56e 100644
--- a/problem-5/problem-5.test.js
+++ b/problem-5/problem-5.test.js
@@ -1,4 +1,27 @@
-const solution = () => {
+/*
+const solution = (num1, num2) => {
+  let smallerNumber = num1;
+  let largerNumber = num2;
+
+  while (true) {
+    const remainder = largerNumber % smallerNumber;
+
+    if (remainder === 0) {
+      return smallerNumber;
+    }
+
+    largerNumber = smallerNumber;
+    smallerNumber = remainder;
+  }
+};
+*/
+
+const solution = (smallerNumber, largerNumber, remainder = largerNumber % smallerNumber) => {
+  if (remainder === 0) {
+    return smallerNumber;
+  }
+
+  return solution(remainder, smallerNumber);
 };
 
 test('최대 공약수를 반환한다', () => {
diff --git a/problem-6/problem-6.test.js b/problem-6/problem-6.test.js
index 059e2b8..c36f19e 100644
--- a/problem-6/problem-6.test.js
+++ b/problem-6/problem-6.test.js
@@ -1,4 +1,47 @@
-const solution = (n) => {
+// const solution = (n) => {
+//  if (n < 0) {
+//    return 0;
+//  }
+//
+//  if (n === 0 || n === 1) {
+//    return 1;
+//  }
+//
+//  return solution(n - 3) + solution(n - 2) + solution(n - 1);
+// };
+
+// 하향식 다이내믹 프로그래밍 (메모이제이션)
+
+// const solution = (n, memo = []) => {
+//  if (n < 0) {
+//    return 0;
+//  }
+//  if (n === 0 || n === 1) {
+//    return 1;
+//  }
+//
+//  if (!memo[n]) {
+//    memo[n] = solution(n - 3, memo) + solution(n - 2, memo) + solution(n - 1, memo);
+//  }
+//
+//  return memo[n];
+// };
+
+// 상향식 다이내믹 프로그래밍
+const solution = (n, current = 2, a = 0, b = 1, c = 1) => {
+  if (n < 0) {
+    return 0;
+  }
+
+  if (n === 0 || n === 1) {
+    return 1;
+  }
+
+  if (n === current) {
+    return a + b + c;
+  }
+
+  return solution(n, current + 1, b, c, a + b + c);
 };
 
 test('계단에 오를 수 있는 가지 수를 반환한다', () => {