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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 numbers are below 100 and multiples of 5 and 10 and 3?
30, 60, 90
How many 5-digit numbers are multiples of 5 and 8?
1
How many multiples of 7 are odd numbers?
odd multiples of 7 are odd numbers.. like 7*1, 7*3,7*5..
How many 3 digit counting numbers are not multiples of 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.
What are all the multiples of 100?
There are an infinite number of multiples of 100. 100, 200, 300,400, ....
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.
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
Are 1 3 5 and 15 a multiples of 15?
No, they are factors of 15. Factors go into numbers, numbers go into multiples.
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 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.
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.
