diff --git a/bona1122/[week8]BackTracking/Archery_competition.js b/bona1122/[week8]BackTracking/Archery_competition.js
new file mode 100644
index 0000000..f8bd20d
--- /dev/null
+++ b/bona1122/[week8]BackTracking/Archery_competition.js
@@ -0,0 +1,64 @@
+// https://school.programmers.co.kr/learn/courses/30/lessons/92342
+// 핵심은 1. 각 점수에 대해 이길지 말지를 결정하는 것
+// 2. 이길 때, 최소한의 화살 사용하기 - 어피치보다 1개만 더 쏘면됨. => 최고점수 만들기 위해 화살 효율적 사용, 그리고 낮은점수를 많이 맞추는게 우선되므로
+
+function solution(n, info) {
+  let maxDiff = 0
+  let answer = [-1]
+  let lion = Array(11).fill(0)
+
+  // level은 현재 검사 중인 점수(0부터 시작), count는 남은 화살 수
+  function dfs(level, count) {
+    if (level == 10) {
+      lion[level] = count // 남은 화살은 모두 0점에 사용
+
+      let sum = 0
+      for (let i = 0; i < 10; i++) {
+        if (lion[i] > info[i]) {
+          // 라이언이 점수 획득
+          sum = sum + (10 - i)
+        } else if (lion[i] === info[i] && lion[i] === 0) {
+          // 둘다 점수획득 못함
+          continue
+        } else if (lion[i] === info[i] || lion[i] < info[i]) {
+          // 어피치가 점수 획득
+          if (info[i] > 0) sum = sum - (10 - i)
+        }
+      }
+
+      if (sum > maxDiff) {
+        maxDiff = sum
+        answer = [...lion]
+      } else if (sum == maxDiff && sum > 0) {
+        // 점수차가 같으면 낮은 점수를 더 많이 맞힌 경우를 선택
+        // 낮은 점수부터 비교
+        for (let j = 10; j >= 0; j--) {
+          if (answer[j] == lion[j]) {
+            continue
+          } else if (lion[j] > answer[j]) {
+            answer = [...lion]
+            break
+          } else {
+            break
+          }
+        }
+      }
+      return
+    }
+
+    // 1. 현재 점수는 포기하는 경우 (화살 0개 사용)
+    dfs(level + 1, count)
+
+    // 2. 현재 점수에서 이기는 경우 (남은화살 있고, 어피치보다 1개 더 사용)
+    const neededArrows = info[level] + 1
+    if (count >= neededArrows) {
+      lion[level] = neededArrows
+      dfs(level + 1, count - neededArrows)
+      lion[level] = 0 // 백트래킹 초기화
+    }
+  }
+
+  dfs(0, n)
+
+  return maxDiff <= 0 ? [-1] : answer
+}
diff --git a/bona1122/[week8]BackTracking/Lotto/Lotto.js b/bona1122/[week8]BackTracking/Lotto/Lotto.js
new file mode 100644
index 0000000..27b423c
--- /dev/null
+++ b/bona1122/[week8]BackTracking/Lotto/Lotto.js
@@ -0,0 +1,31 @@
+// https://www.acmicpc.net/problem/6603
+
+const filePath =
+  process.platform === "linux"
+    ? "/dev/stdin"
+    : require("path").join(__dirname, "input.txt")
+const input = require("fs").readFileSync(filePath).toString().split("\n")
+const log = console.log
+
+// k개의 수를 골라 집합 만들고 그 수만 가지고 번호 선택하기
+let answer = ""
+for (let i = 0; i < input.length; i++) {
+  let [k, ...arr] = input[i].split(" ").map(Number)
+  if (k === 0) break
+
+  const selected = Array(6)
+  const result = []
+  const dfs = (depth, start) => {
+    if (depth === 6) {
+      result.push(selected.join(" "))
+      return
+    }
+    for (let i = start; i < arr.length; i++) {
+      selected[depth] = arr[i]
+      dfs(depth + 1, i + 1)
+    }
+  }
+  dfs(0, 0)
+  answer += result.join("\n") + "\n\n"
+}
+log(answer.trim())
diff --git a/bona1122/[week8]BackTracking/Lotto/input.txt b/bona1122/[week8]BackTracking/Lotto/input.txt
new file mode 100644
index 0000000..f0067af
--- /dev/null
+++ b/bona1122/[week8]BackTracking/Lotto/input.txt
@@ -0,0 +1,3 @@
+7 1 2 3 4 5 6 7
+8 1 2 3 5 8 13 21 34
+0
\ No newline at end of file
diff --git a/bona1122/[week8]BackTracking/NM1/NM1.js b/bona1122/[week8]BackTracking/NM1/NM1.js
new file mode 100644
index 0000000..b54d30b
--- /dev/null
+++ b/bona1122/[week8]BackTracking/NM1/NM1.js
@@ -0,0 +1,28 @@
+// https://www.acmicpc.net/problem/15649
+
+const filePath =
+  process.platform === "linux"
+    ? "/dev/stdin"
+    : require("path").join(__dirname, "input.txt")
+const input = require("fs").readFileSync(filePath).toString()
+const log = console.log
+
+// "순서"가 있게, 1-N 중 "중복없이" M개를 선택하는 문제
+const [N, M] = input.split(" ").map(Number)
+const selected = Array(M) // 총 M개를 고를 것임
+const visited = Array(N + 1).fill(false) // 중복선택 방지 위해
+const dfs = (depth) => {
+  if (depth === M) {
+    log(selected.join(" "))
+    return
+  }
+  for (let i = 1; i <= N; i++) {
+    if (visited[i]) continue
+    selected[depth] = i
+    visited[i] = true
+    dfs(depth + 1)
+    visited[i] = false
+  }
+}
+
+dfs(0, 1)
diff --git a/bona1122/[week8]BackTracking/NM1/input.txt b/bona1122/[week8]BackTracking/NM1/input.txt
new file mode 100644
index 0000000..af1e489
--- /dev/null
+++ b/bona1122/[week8]BackTracking/NM1/input.txt
@@ -0,0 +1 @@
+4 2
\ No newline at end of file
diff --git a/bona1122/[week8]BackTracking/NM2/NM2.js b/bona1122/[week8]BackTracking/NM2/NM2.js
new file mode 100644
index 0000000..c782463
--- /dev/null
+++ b/bona1122/[week8]BackTracking/NM2/NM2.js
@@ -0,0 +1,26 @@
+// https://www.acmicpc.net/problem/15650
+
+const filePath =
+  process.platform === "linux"
+    ? "/dev/stdin"
+    : require("path").join(__dirname, "input.txt")
+const input = require("fs").readFileSync(filePath).toString()
+const log = console.log
+
+// 1-N 중 "중복없이" M개를 선택하는 문제, "오름차순" 정렬 필요
+// => 오름차순 조건으로 인해 for문에서 start 인덱스(현재 선택한 수보다 큰 수부터)를 사용하여 구현
+const [N, M] = input.split(" ").map(Number)
+
+const selected = Array(M)
+const dfs = (depth, start) => {
+  if (depth === M) {
+    log(selected.join(" "))
+    return
+  }
+  for (let i = start; i <= N; i++) {
+    selected[depth] = i
+    dfs(depth + 1, i + 1)
+  }
+}
+
+dfs(0, 1)
diff --git a/bona1122/[week8]BackTracking/NM2/input.txt b/bona1122/[week8]BackTracking/NM2/input.txt
new file mode 100644
index 0000000..af1e489
--- /dev/null
+++ b/bona1122/[week8]BackTracking/NM2/input.txt
@@ -0,0 +1 @@
+4 2
\ No newline at end of file
diff --git a/bona1122/[week8]BackTracking/NM3/NM3.js b/bona1122/[week8]BackTracking/NM3/NM3.js
new file mode 100644
index 0000000..6f564b5
--- /dev/null
+++ b/bona1122/[week8]BackTracking/NM3/NM3.js
@@ -0,0 +1,28 @@
+// https://www.acmicpc.net/problem/15651
+
+const filePath =
+  process.platform === "linux"
+    ? "/dev/stdin"
+    : require("path").join(__dirname, "input.txt")
+const input = require("fs").readFileSync(filePath).toString()
+const log = console.log
+
+// 1-N 중 "중복가능하게" "순서" 있이, M개를 선택하는 문제
+const [N, M] = input.split(" ").map(Number)
+let result = []
+
+const selected = Array(M)
+const dfs = (depth) => {
+  if (depth === M) {
+    result.push(selected.join(" "))
+    return
+  }
+
+  for (let i = 1; i <= N; i++) {
+    selected[depth] = i
+    dfs(depth + 1)
+  }
+}
+
+dfs(0)
+log(result.join("\n"))
diff --git a/bona1122/[week8]BackTracking/NM3/input.txt b/bona1122/[week8]BackTracking/NM3/input.txt
new file mode 100644
index 0000000..af1e489
--- /dev/null
+++ b/bona1122/[week8]BackTracking/NM3/input.txt
@@ -0,0 +1 @@
+4 2
\ No newline at end of file
diff --git a/bona1122/[week8]BackTracking/NM4/NM4.js b/bona1122/[week8]BackTracking/NM4/NM4.js
new file mode 100644
index 0000000..9cb3cd1
--- /dev/null
+++ b/bona1122/[week8]BackTracking/NM4/NM4.js
@@ -0,0 +1,29 @@
+// https://www.acmicpc.net/problem/15652
+
+const filePath =
+  process.platform === "linux"
+    ? "/dev/stdin"
+    : require("path").join(__dirname, "input.txt")
+const input = require("fs").readFileSync(filePath).toString()
+const log = console.log
+
+// 1-N 중 "중복가능하게" M개를 선택하는 문제, "오름차순" 정렬 필요
+// => 오름차순 조건으로 인해 for문에서 start 인덱스(현재 선택한 수 이상)를 사용하여 구현
+
+const [N, M] = input.split(" ").map(Number)
+const selected = Array(M)
+let result = []
+
+const dfs = (depth, start) => {
+  if (depth === M) {
+    result.push(selected.join(" "))
+    return
+  }
+  for (let i = start; i <= N; i++) {
+    selected[depth] = i
+    dfs(depth + 1, i)
+  }
+}
+
+dfs(0, 1)
+log(result.join("\n"))
diff --git a/bona1122/[week8]BackTracking/NM4/input.txt b/bona1122/[week8]BackTracking/NM4/input.txt
new file mode 100644
index 0000000..af1e489
--- /dev/null
+++ b/bona1122/[week8]BackTracking/NM4/input.txt
@@ -0,0 +1 @@
+4 2
\ No newline at end of file
diff --git a/bona1122/[week8]BackTracking/N_Queen.js b/bona1122/[week8]BackTracking/N_Queen.js
new file mode 100644
index 0000000..67a082b
--- /dev/null
+++ b/bona1122/[week8]BackTracking/N_Queen.js
@@ -0,0 +1,30 @@
+// https://school.programmers.co.kr/learn/courses/30/lessons/12952
+
+function solution(n) {
+  let queens = []
+  const possible = (row, col) => {
+    for (let [queenRow, queenCol] of queens) {
+      // 이미 놓인 퀸들과 비교하며, 배치가 가능한지.
+      if (queenRow === row || queenCol === col) return false // 같은 행/열 불가
+      if (Math.abs(queenRow - row) === Math.abs(queenCol - col)) return false // 대각선 불가
+    }
+    return true
+  }
+
+  let cnt = 0
+  const dfs = (row) => {
+    if (row === n) {
+      cnt++
+      return
+    }
+    for (let col = 0; col < n; col++) {
+      if (possible(row, col)) {
+        queens.push([row, col])
+        dfs(row + 1)
+        queens.pop()
+      }
+    }
+  }
+  dfs(0)
+  return cnt
+}
diff --git a/bona1122/[week8]BackTracking/Operator_insertion/Operator_insertion.js b/bona1122/[week8]BackTracking/Operator_insertion/Operator_insertion.js
new file mode 100644
index 0000000..59c4635
--- /dev/null
+++ b/bona1122/[week8]BackTracking/Operator_insertion/Operator_insertion.js
@@ -0,0 +1,49 @@
+// https://www.acmicpc.net/problem/14888
+
+const filePath =
+  process.platform === "linux"
+    ? "/dev/stdin"
+    : require("path").join(__dirname, "input.txt")
+const input = require("fs").readFileSync(filePath).toString().trim().split("\n")
+const log = console.log
+
+// N개로 이루어진 수열, N-1개의 연산자
+// +, -, x, /. (나눗셈은 몫만 취하기, 음수를 양수로 나눌떄는 양수로바꾸어 몫취하고 음수로 바꾸기)
+// 최대, 최소 구하기
+
+const N = +input[0]
+const arr = input[1].split(" ").map(Number)
+const ops = input[2].split(" ").map(Number)
+let max = -Infinity
+let min = Infinity
+
+const dfs = (depth, acc) => {
+  if (depth === N - 1) {
+    max = Math.max(max, acc)
+    min = Math.min(min, acc)
+    return
+  }
+  for (let i = 0; i < 4; i++) {
+    if (ops[i] > 0) {
+      ops[i] -= 1
+      if (i === 0) {
+        dfs(depth + 1, acc + arr[depth + 1])
+      } else if (i === 1) {
+        dfs(depth + 1, acc - arr[depth + 1])
+      } else if (i === 2) {
+        dfs(depth + 1, acc * arr[depth + 1])
+      } else {
+        if (acc < 0) {
+          dfs(depth + 1, -Math.floor(-acc / arr[depth + 1]))
+        } else {
+          dfs(depth + 1, Math.floor(acc / arr[depth + 1]))
+        }
+      }
+      ops[i] += 1
+    }
+  }
+}
+
+dfs(0, arr[0])
+
+log(max + "\n" + min)
diff --git a/bona1122/[week8]BackTracking/Operator_insertion/input.txt b/bona1122/[week8]BackTracking/Operator_insertion/input.txt
new file mode 100644
index 0000000..f2aef3f
--- /dev/null
+++ b/bona1122/[week8]BackTracking/Operator_insertion/input.txt
@@ -0,0 +1,3 @@
+6
+1 2 3 4 5 6
+2 1 1 1
\ No newline at end of file
diff --git a/bona1122/[week8]BackTracking/Prime_finding.js b/bona1122/[week8]BackTracking/Prime_finding.js
new file mode 100644
index 0000000..9e44740
--- /dev/null
+++ b/bona1122/[week8]BackTracking/Prime_finding.js
@@ -0,0 +1,36 @@
+// 소수인지판별하는 함수
+function isPrime(n) {
+  if (n < 2) return false
+  if (n === 2) return true // 2는 소수
+  if (n % 2 === 0) return false // 짝수는 소수가 아님
+
+  for (let i = 3; i <= Math.sqrt(n); i += 2) {
+    if (n % i === 0) return false
+  }
+  return true
+}
+
+// 조각 조합해서 소수 몇개 만들수 있는지 -> "중복없이" "순서있게" 고르기
+function solution(numbers) {
+  const set = new Set()
+  const visited = Array(numbers.length).fill(false)
+  const selected = []
+
+  const dfs = () => {
+    if (selected.length > 0) {
+      const num = Number(selected.join(""))
+      if (isPrime(num)) set.add(num)
+    }
+
+    for (let i = 0; i < numbers.length; i++) {
+      if (visited[i]) continue
+      visited[i] = true
+      selected.push(+numbers[i])
+      dfs()
+      visited[i] = false
+      selected.pop()
+    }
+  }
+  dfs()
+  return set.size
+}
diff --git a/bona1122/[week8]BackTracking/Subsequence_Sum/Subsequence_sum.js b/bona1122/[week8]BackTracking/Subsequence_Sum/Subsequence_sum.js
new file mode 100644
index 0000000..09a6d13
--- /dev/null
+++ b/bona1122/[week8]BackTracking/Subsequence_Sum/Subsequence_sum.js
@@ -0,0 +1,47 @@
+// https://www.acmicpc.net/problem/1182
+
+const filePath =
+  process.platform === "linux"
+    ? "/dev/stdin"
+    : require("path").join(__dirname, "input.txt")
+const input = require("fs").readFileSync(filePath).toString().split("\n")
+const log = console.log
+
+// 크기가 양수인 부분수열 중, 원소를 다 더한 값이 S가 되는 경우의 수 구하기
+const [N, S] = input[0].split(" ").map(Number)
+const arr = input[1].split(" ").map(Number)
+let count = 0
+
+// 1. 비트마스킹을 활용한 부분집합 완전탐색
+// i = 1 부터해서, 크기가 0인 부분집합 제외
+for (let i = 1; i < 1 << N; i++) {
+  let sum = 0
+  for (let j = 0; j < N; j++) {
+    // j번째 원소가 부분집합에 포함되어 있다면
+    if (i & (1 << j)) sum += arr[j]
+  }
+  if (sum === S) count++
+}
+// log(count)
+
+// 2. 백트래킹을 활용한 부분집합 완전탐색
+const selected = Array(N).fill(false)
+let cnt = 0
+const dfs = (depth) => {
+  if (depth === N) {
+    let sum = 0
+    for (let i = 0; i < selected.length; i++) {
+      if (selected[i]) sum += arr[i]
+    }
+    if (sum === S) cnt++
+    return
+  }
+  dfs(depth + 1) // 선택안하고 넘어가기
+
+  // 선택하고 넘어가기
+  selected[depth] = true
+  dfs(depth + 1)
+  selected[depth] = false
+}
+dfs(0)
+log(S === 0 ? --cnt : cnt)
diff --git a/bona1122/[week8]BackTracking/Subsequence_Sum/input.txt b/bona1122/[week8]BackTracking/Subsequence_Sum/input.txt
new file mode 100644
index 0000000..1f5df2c
--- /dev/null
+++ b/bona1122/[week8]BackTracking/Subsequence_Sum/input.txt
@@ -0,0 +1,2 @@
+5 0
+-7 -3 -2 5 8
\ No newline at end of file