# Calculate the number of 1,500th divisible by 2, 3, or 5.

Source: Internet
Author: User

Method 1: brute-force enumeration 2, 3, 4..., Judge in turn whether N can be divided by 2 | 3 | 5 until 1,500th

`#include <iostream>//#include <string.h>using namespace std;int main(int argc, char *argv[]){    int count = 1;    int i;    for(i=2;count!=1500;i++)    {            cout<<i<<endl;            if(i%2==1 || i%3==1 || i%5==1)                      count++;                }    cout<<i<<endl;    getchar();    return 0;}`

Let's talk a little bit about the skills. The running speed is similar.

If X can be divisible by 2, 3, or 5, then X must be in the form of 2y, 3Y, or 5y.

Therefore, we only need to list the numbers of 2y, 3Y, and 5y in ascending order until we can list them into 1500, even if we want to solve them.

Here, there may be two or three expressions satisfying at the same time.

`#include<iostream>#include<stack>using namespace std;int max(int a,int b,int c){    if(a > b)    {         if(a > c)         {              return a;         }else{               return c;         }    }else    {         if(b>c)         {                return b;         }else{               return c;         }     }}int function(int a,int b,int c,int n){    int count = 1;    int maxa = a;    int maxb = b;    int maxc = c;     while(count<n)    {         cout<<count<<maxa<<maxb<<maxc<<endl;              if(maxa < maxb && maxa <maxc)              {                     count++;                     maxa += a;               }                 else if(maxb < maxa && maxb <maxc)              {                     count++;                     maxb += b;               }              else if(maxc < maxb && maxc <maxa)              {                     count++;                     maxc += c;               }else if(maxa == maxb && maxc > maxa)              {                    count++;                    maxa += a;                    maxb += b;              }              else if(maxc == maxb && maxa > maxc)              {                    count++;                    maxc += c;                    maxb += b;              }              else if(maxa == maxc && maxb > maxa)              {                    count++;                    maxa += a;                    maxc += c;              }else{                    count++;                    maxa += a;                    maxb += b;                    maxc += c;              }                      }        return max(maxa,maxb,maxc);}int main(){    cout<<function(2,3,5,1500);    getchar();    getchar();    return 0;}`

I remember another way: the numbers that meet the conditions within 30 are: {2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30} total 22

So find the number of n that meets the condition, first obtain the number (N/22) * 22 (n is an integer), is (N/22) * 30, then, start from this number and calculate n % 22 numbers that meet the conditions.

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