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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 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 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 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.


What are multiples of 5 between 1 to 100?

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


How do you use a sieve of Eratosthenes?

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.


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.