试图找到可被1到20的所有数字均分的最小正数

Question

I'm trying to find the smallest positive number that is evenly divisible by all of the numbers from 1 to 20. We are given that 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. My find() finds the number starting from 2520 that is divisible by all numbers from 1-20 but is returning 2520 for some reason. I cannot find what's wrong about my find()?

  public class Solution  {

  public ArrayList<Integer> list = new ArrayList<Integer>();


// creating a list of integers from 1 to 20
public ArrayList<Integer> addtolist() {
  for (int i = 1; i <= 20; i++) {
    list.add(i);
  }
  return list;
}

// finds the smallest positive number that is evenly divisible by all 
of the numbers from 1 to 20

public int find() {
  int num = 2520;
  while(true) {
    for(int i: list) {
      if(num % i == 0) {
        return num;
      }
      else {
        num = num + 1;
      }

    }
  }
}

public static void main(String[] args) {
  Solution sol = new Solution();
  sol.addtolist();
  System.out.println(sol.find());//2520
}


}

Answer 1

Your find function returns num if any i in list divides it. It should only return num if every i in num is a divisor.

Although it has to be said that this is far from the most efficient solution to the problem.

Answer 2

You return from the for loop when (num % i == 0) , given that i starts at 1 this always true. Instead you need to wait until the end to return:

public int find() {
  int num = 2520;
  while(true) {
    boolean allDivisible = true;
    for(int i: list) {
      if(num % i != 0) {
        allDivisible = false;
        break;
      }
    }
    if (allDivisible) {
      return num;
    else {
      num = num + 1;
    }
  }
}

Answer 3

In your code:

for(int i: list) {
  if(num % i == 0) {
    return num; // Returns immediately. 
  }
  else {
    num = num + 1;
  }
}

you return as soon as you find some number in the list that has a match. What you want to do is only return when you have found a value that matches all in the list.

Answer 4

Nice question!

long answer = LongStream.iterate(1, n -> n + 1)
    .filter(n -> IntStream.rangeClosed(1, 20).allMatch(f -> n % f == 0))
    .findFirst().getAsLong();

Answer is 232792560

There are obviously plenty of shortcuts using math (eg only looking at even numbers, ignoring numbers in 1 to 20 that are factors of other numbers in that range).