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5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100.

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What is the ratio of the number of multiples of 5 between 1 to 100 to all numbers 1 to 100?

1 to 5


How do you write a C program using an integer array to print array numbers up to 100 and then remove all numbers that are divisible by 2 and print out the remaining numbers and continue for 3 to 9?

#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(); }


What percent of all numbers between 1 to 100 are multiples of 5?

20 percent.


What is the ratio of the number of positive common multiples less than 100 of the numbers 3 and 5 to the number of positive common multiples less than 100 of the numbers 2 and 7?

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.


How many 3 digit numbers are multiples of 5?

To determine the number of 3-digit numbers that are multiples of 5, we need to find the first and last 3-digit multiples of 5. The first 3-digit multiple of 5 is 100, and the last 3-digit multiple of 5 is 995. To find the total number of such multiples, we can use the formula (Last - First) / 5 + 1 = (995 - 100) / 5 + 1 = 180. Therefore, there are 180 3-digit numbers that are multiples of 5.


Which of the multiples of 5 that are less than 100 are also multiples of 6?

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


What is The first 5 multiples?

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.


What are the numbers that are divisible by 5 in the range 1-100?

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.


What are multiples of 5 between 1 to 100?

5,10,15,20,25,30,35,40,45,50,55,60


What numbers under 100 have at least a 2 or 5 in their prime factorization?

All the even numbers and all the odd multiples of 5.


What are the 100 multiples of 5?

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.


Are 1 3 5 and 15 a multiples of 15?

No, they are factors of 15. Factors go into numbers, numbers go into multiples.