what is the ratio of multiples of 3 between 1 and 100 to all numbers between 1 and 100
25%
Numbers from 0 to 100 that are multiples of 10 are: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100. That's ten numbers, so the fraction of numbers from 1 to 100 that are multiples of 10 is 10/100. In simplest form it is 1/10.
All of those numbers have an infinite amount of multiples.
The only even prime number (not only from 1 to 100, but in general) is 2. All other even numbers are multiples of 2, and therefore, by definition, not prime numbers.
what is the ratio of multiples of 3 between 1 and 100 to all numbers between 1 and 100
All of those numbers have an infinite number of multiples.
All numbers have an infinite amount of multiples.
101
All of them.
All numbers have an infinite amount of multiples.
All of those numbers have infinite multiples.
33 to 100
1 to 5
25%
Multiples of 100.
#include<stdio.h> #include<conio.h> void main() { int i, a[100]; clrscr(); printf("Numbers from 1 to 100"); for(i=1;i<=100;i++) printf(" %d",a[i]); printf("Numbers from 1 to 100 without multiples of 2"); for(i=1;i<=100;i++) { if (a[i]%2 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 3"); for(i=1;i<=100;i++) { if (a[i]%3 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 4"); for(i=1;i<=100;i++) { if (a[i]%4 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 5"); for(i=1;i<=100;i++) { if (a[i]%5 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 6"); for(i=1;i<=100;i++) { if (a[i]%6 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 7"); for(i=1;i<=100;i++) { if (a[i]%7 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 8"); for(i=1;i<=100;i++) { if (a[i]%8 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 9"); for(i=1;i<=100;i++) { if (a[i]%9 !=0) printf(" %d",a[i]); } getch(); }