diff --git a/Challenge/jmjmjm/everyArray/README.md b/Challenge/jmjmjm/everyArray/README.md
new file mode 100644
index 00000000..442558f6
--- /dev/null
+++ b/Challenge/jmjmjm/everyArray/README.md
@@ -0,0 +1,9 @@
+# 문제제목
+
+## 설명
+
+every를 이용해서 모든 원소가 짝수인지 아닌지를 판별하세요
+
+## Expected Output 
+
+true
\ No newline at end of file
diff --git a/Challenge/jmjmjm/everyArray/solve.js b/Challenge/jmjmjm/everyArray/solve.js
new file mode 100644
index 00000000..9b97d897
--- /dev/null
+++ b/Challenge/jmjmjm/everyArray/solve.js
@@ -0,0 +1,8 @@
+const inputArray = [2, 3, 6, 8, 10];
+
+let answer = inputArray.every(item => {
+    return item % 2 == 0;
+})
+
+console.log(answer);
+
diff --git a/Challenge/jmjmjm/expDivOdd/README.md b/Challenge/jmjmjm/expDivOdd/README.md
new file mode 100644
index 00000000..c07ab10b
--- /dev/null
+++ b/Challenge/jmjmjm/expDivOdd/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+제곱한 후 3으로 나눈 나머지가 홀수인 것 을 뽑은 배열의 총 합을 구하세요.
diff --git a/Challenge/jmjmjm/expDivOdd/solve.js b/Challenge/jmjmjm/expDivOdd/solve.js
new file mode 100644
index 00000000..2f90df3a
--- /dev/null
+++ b/Challenge/jmjmjm/expDivOdd/solve.js
@@ -0,0 +1,12 @@
+// 제곱한 후 3으로 나눈 나머지가 홀수인 것 을 뽑은 배열의 총 합을 구하세요.
+
+const inputArray = [2, 3, 6, 8, 10];
+
+let answer = inputArray
+  .map((e) => e ** 2)
+  .filter((e) => e % 3 == 1)
+  .reduce((acc, cur) => {
+    return acc + cur;
+  }, 0);
+
+console.log(answer);
diff --git a/Challenge/jmjmjm/figureSkating/README.md b/Challenge/jmjmjm/figureSkating/README.md
new file mode 100644
index 00000000..54b85fbd
--- /dev/null
+++ b/Challenge/jmjmjm/figureSkating/README.md
@@ -0,0 +1,35 @@
+## 설명
+
+점수를 계산해서 점수가 높은 순으로 name과 score를 가진 객체의 배열을 출력하세요.
+단 실격자는 출력하지 않습니다.
+
+### 피겨 점수 기준
+
+피겨 점수는 ‘기술점수(TES)+예술점수(PCS)-감점’으로 구성된다.
+
+기술점수(TES·Total Element Score)는 기본점수에 수행점수(GOE·Grade Of Execution)를 합산해 도출한다.
+
+심판은 선수들이 점프와 스핀의 기술을 제대로 수행하는지 살핀다. 점프에선 회전수를 제대로 지켰는지, 에지를 제대로 사용했는지에 따라 ‘롱 에지’(잘못된 에지 사용)나 ‘다운그레이드’(난이도 하락), ‘어텐션’(주의) 등의 판정을 내린다. 스핀과 스텝시퀀스에 붙는 레벨(1~4)도 이들이 결정한다.
+
+9명의 심판은 선수들이 미리 제출한 연기 구성표를 기준으로 과제별 기본점수에서 가·감점을 한다. 이른바 ‘GOE’라 불리는 수행점수다.
+
+쇼트프로그램은 점프 3개, 스핀 3개, 스텝 1개의 수행과제가 반드시 포함돼야 한다.
+
+예술점수(PCS·Total Program Component Score)는 프로그램의 완성도에 영향을 미치는 스케이팅 기술·동작의 연결·연기·안무·곡 해석 등 5가지를 평가한다.
+
+마지막으로 감점이 반영된다. 감점 항목이 발생할 때마다 1점씩 감점된다.
+
+원문보기:
+http://news.khan.co.kr/kh_news/khan_art_view.html?art_id=201402172146545#csidx6329aebb6a02152bca884614a7f0544
+
+### 판정 기준
+
+- 점수가 가장 높은 사람이 1등이다.
+- 점수는 기술점수(TES) + 예술점수(PCS) - 감점 으로 구성된다.
+- 기술점수(TES)는 기본점수 \* 수행점수(GOE)를 합산해 도출한다.
+- 필수 수행과제를 수행하지 못했을 시 실격이다.
+
+### 필수 수행과제
+
+- 쇼트프로그램은 점프 3개, 스핀 3개, 스텝 1개의 수행과제가 반드시 포함돼야 한다.
+- 점프와 스핀을 4번 이상 수행했을 시 가장 잘한 3개를 기준으로 한다.
diff --git a/Challenge/jmjmjm/figureSkating/solve.js b/Challenge/jmjmjm/figureSkating/solve.js
new file mode 100644
index 00000000..3b16aa19
--- /dev/null
+++ b/Challenge/jmjmjm/figureSkating/solve.js
@@ -0,0 +1,87 @@
+// 기본 점수
+const baseScore = {
+  jump: 10,
+  spin: 20,
+  step: 15,
+};
+
+const baseAssignment = {
+  jump: 3,
+  spin: 3,
+  step: 1,
+};
+
+const inputArray = [
+  {
+    name: "김연아",
+    goe: {
+      jump: [10, 9, 10],
+      spin: [10, 10, 10],
+      step: [9],
+    },
+    pcs: 30,
+    penalty: 0,
+  },
+  {
+    name: "피겨의 신",
+    goe: {
+      jump: [10000, 10000],
+      spin: [10000, 10000, 10000],
+      step: [10000],
+    },
+    pcs: 100000,
+    penalty: 0,
+  },
+  {
+    name: "아사다 마오",
+    goe: {
+      jump: [9, 6, 8, 9],
+      spin: [9, 9, 9],
+      step: [8],
+    },
+    pcs: 29,
+    penalty: 10,
+  },
+];
+
+const answer = inputArray
+  .filter((player) => {
+    return (
+      player.goe.jump.length >= baseAssignment.jump &&
+      player.goe.spin.length >= baseAssignment.spin &&
+      player.goe.step.length >= baseAssignment.step
+    );
+  })
+  .map((v) => {
+    //TES
+    let result = 0;
+    result += v.goe.jump
+      .sort((a, b) => b - a)
+      .slice(0, baseAssignment.jump)
+      .reduce((acc, cur) => {
+        return acc + cur * baseScore.jump;
+      }, 0);
+    result += v.goe.spin
+      .sort((a, b) => b - a)
+      .slice(0, baseAssignment.spin)
+      .reduce((acc, cur) => {
+        return acc + cur * baseAssignment.spin;
+      }, 0);
+    result += v.goe.step
+      .sort((a, b) => b - a)
+      .slice(0, baseAssignment.step)
+      .reduce((acc, cur) => {
+        return acc + cur * baseAssignment.step;
+      }, 0);
+    // PCS
+    result += v.pcs;
+    // PENALTY
+    result -= v.penalty;
+
+    return {
+      name: v.name,
+      score: result,
+    };
+  });
+
+console.log(answer);
diff --git a/Challenge/jmjmjm/filterAge/README.md b/Challenge/jmjmjm/filterAge/README.md
new file mode 100644
index 00000000..873a15c1
--- /dev/null
+++ b/Challenge/jmjmjm/filterAge/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+배열 원소의 age가 30이상 50미만인 사람만 있는 객체의 배열을 만드세요
diff --git a/Challenge/jmjmjm/filterAge/solve.js b/Challenge/jmjmjm/filterAge/solve.js
new file mode 100644
index 00000000..6305168c
--- /dev/null
+++ b/Challenge/jmjmjm/filterAge/solve.js
@@ -0,0 +1,19 @@
+// 배열 원소의 age가 30이상 50미만인 사람만 있는 객체의 배열을 만드세요
+// write your codes
+
+const inputArray = [
+    {name: '재만', age: 36},
+    {name: '준호', age: 30},
+    {name: '영희', age: 28},
+    {name: '은혜', age: 45},
+    {name: '영수', age: 55}
+];
+
+let answer = inputArray.filter(e => e.age >= 30 && e.age < 50)
+        .sort(function(a, b) {
+            if (a.age > b.age) return 1;
+            if (a.age == b.age) return 0;
+            if (a.age < b.age) return -1;
+        })
+
+console.log(answer);
diff --git a/Challenge/jmjmjm/filterIntersection/README.md b/Challenge/jmjmjm/filterIntersection/README.md
new file mode 100644
index 00000000..d7c9ad39
--- /dev/null
+++ b/Challenge/jmjmjm/filterIntersection/README.md
@@ -0,0 +1,5 @@
+# 문제제목
+
+## 설명
+
+두 배열의 교집합을 출력하세요!
diff --git a/Challenge/jmjmjm/filterIntersection/solve.js b/Challenge/jmjmjm/filterIntersection/solve.js
new file mode 100644
index 00000000..6f07c8a3
--- /dev/null
+++ b/Challenge/jmjmjm/filterIntersection/solve.js
@@ -0,0 +1,8 @@
+const tempA = [1, 2, 3, 4, 5];
+const tempB = [3, 4, 5, 6, 7];
+
+const answer = tempA.filter(e => tempB.includes(e));
+
+console.log(answer);
+
+
diff --git a/Challenge/JM Lim/filterOdd/README.md b/Challenge/jmjmjm/filterOdd/README.md
similarity index 100%
rename from Challenge/JM Lim/filterOdd/README.md
rename to Challenge/jmjmjm/filterOdd/README.md
diff --git a/Challenge/JM Lim/filterOdd/solve.js b/Challenge/jmjmjm/filterOdd/solve.js
similarity index 96%
rename from Challenge/JM Lim/filterOdd/solve.js
rename to Challenge/jmjmjm/filterOdd/solve.js
index a6895962..f8c9d11d 100644
--- a/Challenge/JM Lim/filterOdd/solve.js	
+++ b/Challenge/jmjmjm/filterOdd/solve.js
@@ -2,6 +2,6 @@ const inputArray = [1, 5, 6, 7 ,9, 3, 14];
 
 const answer = inputArray.filter(e => {
     return (e%2) == 1
-})
+});
 
 console.log(answer);
\ No newline at end of file
diff --git a/Challenge/jmjmjm/findWord/README.md b/Challenge/jmjmjm/findWord/README.md
new file mode 100644
index 00000000..8ef123f4
--- /dev/null
+++ b/Challenge/jmjmjm/findWord/README.md
@@ -0,0 +1,9 @@
+# 문제제목
+
+## 설명
+
+용가리라는 단어가 있으면 true 없으면 false를 출력
+
+## Expected Output 
+
+true
diff --git a/Challenge/jmjmjm/findWord/solve.js b/Challenge/jmjmjm/findWord/solve.js
new file mode 100644
index 00000000..e63b8ec5
--- /dev/null
+++ b/Challenge/jmjmjm/findWord/solve.js
@@ -0,0 +1,5 @@
+const inputArray = ["잠", "자", "고", "싶", "다", "용가리", "jmjmjm"];
+
+const answer = inputArray.includes("용가리");
+
+console.log(answer);
diff --git a/Challenge/JM Lim/forEachFilter/README.md b/Challenge/jmjmjm/forEachFilter/README.md
similarity index 100%
rename from Challenge/JM Lim/forEachFilter/README.md
rename to Challenge/jmjmjm/forEachFilter/README.md
diff --git a/Challenge/JM Lim/forEachFilter/solve.js b/Challenge/jmjmjm/forEachFilter/solve.js
similarity index 97%
rename from Challenge/JM Lim/forEachFilter/solve.js
rename to Challenge/jmjmjm/forEachFilter/solve.js
index a8f983fc..43822b1d 100644
--- a/Challenge/JM Lim/forEachFilter/solve.js	
+++ b/Challenge/jmjmjm/forEachFilter/solve.js
@@ -4,6 +4,6 @@ const answer = [];
 
 inputArray.forEach(e => {
     e >= 40 ? answer.push(e) : ''
-})
+});
 
 console.log(answer);
\ No newline at end of file
diff --git a/Challenge/JM Lim/forEachFilterIsNaN/README.md b/Challenge/jmjmjm/forEachFilterIsNaN/README.md
similarity index 100%
rename from Challenge/JM Lim/forEachFilterIsNaN/README.md
rename to Challenge/jmjmjm/forEachFilterIsNaN/README.md
diff --git a/Challenge/JM Lim/forEachFilterIsNaN/solve.js b/Challenge/jmjmjm/forEachFilterIsNaN/solve.js
similarity index 97%
rename from Challenge/JM Lim/forEachFilterIsNaN/solve.js
rename to Challenge/jmjmjm/forEachFilterIsNaN/solve.js
index 72205624..8ba2517d 100644
--- a/Challenge/JM Lim/forEachFilterIsNaN/solve.js	
+++ b/Challenge/jmjmjm/forEachFilterIsNaN/solve.js
@@ -6,6 +6,7 @@ inputArray.forEach(e => {
     if (Number.isInteger(e)) {
         answer.push(e)
     }
-})
+});
+
 console.log(answer);
 
diff --git a/Challenge/JM Lim/forEachMap/README.md b/Challenge/jmjmjm/forEachMap/README.md
similarity index 100%
rename from Challenge/JM Lim/forEachMap/README.md
rename to Challenge/jmjmjm/forEachMap/README.md
diff --git a/Challenge/JM Lim/forEachMap/solve.js b/Challenge/jmjmjm/forEachMap/solve.js
similarity index 97%
rename from Challenge/JM Lim/forEachMap/solve.js
rename to Challenge/jmjmjm/forEachMap/solve.js
index 09fc7b9c..c47424d5 100644
--- a/Challenge/JM Lim/forEachMap/solve.js	
+++ b/Challenge/jmjmjm/forEachMap/solve.js
@@ -5,6 +5,6 @@ const answer = [];
 inputArray.forEach(e => {
     let val = e+'';
     answer.push(val.concat('%'))
-})
+});
 
 console.log(answer);
\ No newline at end of file
diff --git a/Challenge/JM Lim/forEachReduce/README.md b/Challenge/jmjmjm/forEachReduce/README.md
similarity index 100%
rename from Challenge/JM Lim/forEachReduce/README.md
rename to Challenge/jmjmjm/forEachReduce/README.md
diff --git a/Challenge/JM Lim/forEachReduce/solve.js b/Challenge/jmjmjm/forEachReduce/solve.js
similarity index 95%
rename from Challenge/JM Lim/forEachReduce/solve.js
rename to Challenge/jmjmjm/forEachReduce/solve.js
index 41d75ad5..2e49b6c0 100644
--- a/Challenge/JM Lim/forEachReduce/solve.js	
+++ b/Challenge/jmjmjm/forEachReduce/solve.js
@@ -4,6 +4,6 @@ let answer = '';
 
 answer = inputArray.reduce((a, c) => {
     return a+c
-}, 0)
+}, 0);
 
 console.log(answer);
diff --git a/Challenge/jmjmjm/mapAddPercent/README.md b/Challenge/jmjmjm/mapAddPercent/README.md
new file mode 100644
index 00000000..75e73042
--- /dev/null
+++ b/Challenge/jmjmjm/mapAddPercent/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+map 메소드를 사용해 배열 각각 숫자 뒤에 %를 붙인 문자열을 만드세요
diff --git a/Challenge/jmjmjm/mapAddPercent/solve.js b/Challenge/jmjmjm/mapAddPercent/solve.js
new file mode 100644
index 00000000..ac162a0e
--- /dev/null
+++ b/Challenge/jmjmjm/mapAddPercent/solve.js
@@ -0,0 +1,5 @@
+const inputArray = [100, 10, 20, 40];
+
+const answer = inputArray.map((e) => e + "%");
+
+console.log(answer);
diff --git a/Challenge/jmjmjm/mapAppendOrder/README.md b/Challenge/jmjmjm/mapAppendOrder/README.md
new file mode 100644
index 00000000..1435623a
--- /dev/null
+++ b/Challenge/jmjmjm/mapAppendOrder/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+배열의 값을 name 프로퍼티에 넣고 몇번째 원소인지를 order에 넣은 객체의 배열을 출력하세요
diff --git a/Challenge/jmjmjm/mapAppendOrder/solve.js b/Challenge/jmjmjm/mapAppendOrder/solve.js
new file mode 100644
index 00000000..11caf804
--- /dev/null
+++ b/Challenge/jmjmjm/mapAppendOrder/solve.js
@@ -0,0 +1,8 @@
+const inputArray = ["홍길동", "둘리", "루피"];
+
+const answer = inputArray.map((el, idx) => {
+  return {
+    name: el,
+    order: idx + 1,
+  };
+});
diff --git a/Challenge/jmjmjm/recommendAd/README.md b/Challenge/jmjmjm/recommendAd/README.md
new file mode 100644
index 00000000..132e8800
--- /dev/null
+++ b/Challenge/jmjmjm/recommendAd/README.md
@@ -0,0 +1,15 @@
+# 문제제목
+
+사용자가 안 본 광고는?
+
+## 설명
+
+유튜브는 사용자가 프리미엄 회원이 아닌 경우 영상 시작이나 중간에 광고가 나오도록 설정되어 있습니다.
+
+[2020년 유튜브 인기 광고](https://www.thinkwithgoogle.com/intl/ko-kr/marketing-strategies/video/%EB%8C%80%ED%95%9C%EB%AF%BC%EA%B5%AD-youtube-%EC%9D%B8%EA%B8%B0-%EA%B4%91%EA%B3%A0%EC%98%81%EC%83%81-2020%EB%85%84-%EC%97%B0%EB%A7%90%EA%B2%B0%EC%82%B0/)는 총 열 편으로, 사용자의 일주일간 광고 시청 이력을 통해 해당 사용자가 일주일 동안 안 본 광고를 그 다음주에 노출함으로써 광고들이 사용자에게 골고루 노출되길 원합니다.
+
+유저가 매일 유튜브에 접속하여 하루 한 편 이상의 광고를 보았다고 가정할 때, 이 유저가 안 본 광고는 무엇인지 출력해주세요.
+
+## Expected Output
+
+[ '동원F&B' ]
diff --git a/Challenge/jmjmjm/recommendAd/solve.js b/Challenge/jmjmjm/recommendAd/solve.js
new file mode 100644
index 00000000..c4eff6d6
--- /dev/null
+++ b/Challenge/jmjmjm/recommendAd/solve.js
@@ -0,0 +1,28 @@
+const userHistory = [
+  { date: "2020-11-03", watched: ["KT", "BBQ"] },
+  { date: "2020-11-04", watched: ["정관장", "KT", "딱좋은데이"] },
+  { date: "2020-11-05", watched: ["그랑사가", "농심"] },
+  { date: "2020-11-06", watched: ["BBQ"] },
+  { date: "2020-11-07", watched: ["쌍용자동차", "BBQ", "KT"] },
+  { date: "2020-11-08", watched: ["켈로그코리아", "빙그레"] },
+  { date: "2020-11-09", watched: ["KT", "그랑사가", "빙그레"] },
+];
+
+const advList = [
+  "동원F&B",
+  "정관장",
+  "KT",
+  "BBQ",
+  "그랑사가",
+  "농심",
+  "딱좋은데이",
+  "빙그레",
+  "쌍용자동차",
+  "켈로그코리아",
+];
+let allAdv = [];
+
+userHistory.forEach((el) => allAdv.push(...el.watched));
+const answer = advList.filter((e) => !allAdv.includes(e));
+
+console.log(answer);
diff --git a/Challenge/jmjmjm/reduceMaxValueNIndex/README.md b/Challenge/jmjmjm/reduceMaxValueNIndex/README.md
new file mode 100644
index 00000000..965bdec6
--- /dev/null
+++ b/Challenge/jmjmjm/reduceMaxValueNIndex/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+reduce 메소드를 사용해 최댓값의 값을 maxValue에, 해당 값의 index를 idx에 넣은 객체를 출력하세요
diff --git a/Challenge/jmjmjm/reduceMaxValueNIndex/solve.js b/Challenge/jmjmjm/reduceMaxValueNIndex/solve.js
new file mode 100644
index 00000000..36baa9f8
--- /dev/null
+++ b/Challenge/jmjmjm/reduceMaxValueNIndex/solve.js
@@ -0,0 +1,4 @@
+// write your codes
+function solution(inputArray) {}
+
+exports.solution = solution;
diff --git a/Challenge/jmjmjm/reduceNameNickname/README.md b/Challenge/jmjmjm/reduceNameNickname/README.md
new file mode 100644
index 00000000..b6920e05
--- /dev/null
+++ b/Challenge/jmjmjm/reduceNameNickname/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+입력받은 객채배열의 nickname을 key, name을 value로 하는 객체를 출력하세요
diff --git a/Challenge/jmjmjm/reduceNameNickname/solve.js b/Challenge/jmjmjm/reduceNameNickname/solve.js
new file mode 100644
index 00000000..36baa9f8
--- /dev/null
+++ b/Challenge/jmjmjm/reduceNameNickname/solve.js
@@ -0,0 +1,4 @@
+// write your codes
+function solution(inputArray) {}
+
+exports.solution = solution;
diff --git a/Challenge/jmjmjm/reduceSum/README.md b/Challenge/jmjmjm/reduceSum/README.md
new file mode 100644
index 00000000..8c607eb2
--- /dev/null
+++ b/Challenge/jmjmjm/reduceSum/README.md
@@ -0,0 +1,9 @@
+# 문제제목
+
+## 설명
+
+reduce 메소드를 사용해서 배열의 모든 수의 합을 구하세요.
+
+## Expected Output 
+
+106
diff --git a/Challenge/jmjmjm/reduceSum/solve.js b/Challenge/jmjmjm/reduceSum/solve.js
new file mode 100644
index 00000000..36baa9f8
--- /dev/null
+++ b/Challenge/jmjmjm/reduceSum/solve.js
@@ -0,0 +1,4 @@
+// write your codes
+function solution(inputArray) {}
+
+exports.solution = solution;
diff --git a/Challenge/jmjmjm/sortByPrice/README.md b/Challenge/jmjmjm/sortByPrice/README.md
new file mode 100644
index 00000000..b1ccf70e
--- /dev/null
+++ b/Challenge/jmjmjm/sortByPrice/README.md
@@ -0,0 +1,14 @@
+# 문제제목
+
+## 설명
+
+배열안의 객체를 price를 기준으로 오름차순 정렬한 배열을 출력하세요
+
+## Expected Output 
+
+[
+  { name: '사과', price: 1000 },
+  { name: '당근', price: 2000 },
+  { name: '수박', price: 5000 },
+  { name: '참외', price: 10000 }
+]
diff --git a/Challenge/jmjmjm/sortByPrice/solve.js b/Challenge/jmjmjm/sortByPrice/solve.js
new file mode 100644
index 00000000..36baa9f8
--- /dev/null
+++ b/Challenge/jmjmjm/sortByPrice/solve.js
@@ -0,0 +1,4 @@
+// write your codes
+function solution(inputArray) {}
+
+exports.solution = solution;
diff --git a/Challenge/jmjmjm/sortByPriceAndQuantity/README.md b/Challenge/jmjmjm/sortByPriceAndQuantity/README.md
new file mode 100644
index 00000000..41cd5c95
--- /dev/null
+++ b/Challenge/jmjmjm/sortByPriceAndQuantity/README.md
@@ -0,0 +1,16 @@
+# 문제제목
+
+## 설명
+
+배열안의 객체를 price를 기준으로 오름차순 정렬한 배열을 출력하세요
+만약 price가 같다면 quantity기준으로 오름차순 정렬하세요
+
+## Expected Output 
+
+[
+  { name: '사과', price: 1000, quantity: 2 },
+  { name: '오이', price: 2000, quantity: 49 },
+  { name: '당근', price: 2000, quantity: 50 },
+  { name: '참외', price: 5000, quantity: 10 },
+  { name: '수박', price: 5000, quantity: 20 }
+]
diff --git a/Challenge/jmjmjm/sortByPriceAndQuantity/solve.js b/Challenge/jmjmjm/sortByPriceAndQuantity/solve.js
new file mode 100644
index 00000000..36baa9f8
--- /dev/null
+++ b/Challenge/jmjmjm/sortByPriceAndQuantity/solve.js
@@ -0,0 +1,4 @@
+// write your codes
+function solution(inputArray) {}
+
+exports.solution = solution;
diff --git a/Challenge/jmjmjm/spreadOperatorMaxValue/README.md b/Challenge/jmjmjm/spreadOperatorMaxValue/README.md
new file mode 100644
index 00000000..164b68a1
--- /dev/null
+++ b/Challenge/jmjmjm/spreadOperatorMaxValue/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+Spread Operator를 이용해서 값의 최대, 최소값을 구하세요
diff --git a/Challenge/jmjmjm/spreadOperatorMaxValue/solve.js b/Challenge/jmjmjm/spreadOperatorMaxValue/solve.js
new file mode 100644
index 00000000..36baa9f8
--- /dev/null
+++ b/Challenge/jmjmjm/spreadOperatorMaxValue/solve.js
@@ -0,0 +1,4 @@
+// write your codes
+function solution(inputArray) {}
+
+exports.solution = solution;