To determine the factors of 100 that are multiples of 5, we first find the factors of 100, which are 1, 2, 4, 5, 10, 20, 25, 50, and 100. Among these factors, the multiples of 5 are 5, 10, 20, 25, 50, and 100. Therefore, there are 6 factors of 100 that are multiples of 5.
30, 60, 90
2250
odd multiples of 7 are odd numbers.. like 7*1, 7*3,7*5..
There are 720 of them. The three digit counting numbers are 100-999. All multiples of 5 have their last digit as 0 or 5. There are 9 possible numbers {1-9} for the first digit, There are 10 possible numbers {0-9} for each of the first digits, There are 8 possible numbers {1-4, 6-9} for each of the first two digits, Making 9 x 10 x 8 = 720 possible 3 digit counting numbers not multiples of 5.
1 to 5
#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(); }
20 percent.
The LCM of two consecutive numbers is their product. The LCM of two consecutive multiples of 5 is their product divided by 5. Two consecutive numbers cannot be multiples of 5.
The multiples of 5 that are also multiples of 6 are multiples of their LCM. The LCM of 5 and 6 is 30, so the question becomes which multiples of 30 are less than 100? The solution is the numbers: 30, 60, 90
Since you didn't specify a single number, and all numbers are multiples of themselves, the five smallest multiples are the counting numbers 1 to 5.
1 ÷ 5 = 0 r 1 → first multiple of 5 in the range 1-100 is 1 x 5 = 5 100 ÷ 5 = 20 → last multiples of 5 in the range 1-100 is 20 x 5 = 100 → want the first 20 multiples of 5, namely: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100.
5,10,15,20,25,30,35,40,45,50,55,60
They are the hundred numbers of the form 5*k where k = 1, 2, 3, ... So, get a calculator and work out 5 x 1 5 x 2 5 x 3 all the way to 5 x 100.
No, they are factors of 15. Factors go into numbers, numbers go into multiples.
All the even numbers and all the odd multiples of 5.
1.draw up a chart 1-100 2.get rid of 2's multiples 3.get rid of 3's multiples 4.get rid of 5's multiples 5.get rid of 7's multiples 6. all the left over numbers are prime.