You will need some variables first:
amount - store the amount of "divisible by 7" integers
total - store the sum of these integers
count - a number that will control the while loop (or you can use a for loop an exclude this)
amount = 0, total = 0, and count = 101
A special math operator is also needed, it's called modulus, it's the same as division except instead of dividing a number it finds the remainder. It uses a percent sign %
Modulus (or Mod for short) is important for this type of problem, as it can easily tell us if a number is divisible by 7. If the number Mod 7 equals 0 then it means the number is divisible by 7.
Now the while loop:
while (count < 200) {
if (count % 7 == 0) {
amount++;
total += count;
}
count++;
}
When this loop is done, amount will be the amount of numbers that are divisible by 7, and total will be all of those numbers added together. Simply print those two variables using System.out.println(); and you will have your answer.
If you want to print out each number onto the screen, just insert System.out.printlin(count); into the if statement of the while loop, then every number that's divisible by 7 will be shown on screen.
Write a program to find the number and sum of all integers from 100 to 300 that are divisible by 11
if ( x % 6 == 0 ){ printf( "%d is divisible by 6", x ); } else { printf( "%d is not divisible by 6", x ); }
bool is_even(long int num) { return !(num & 1); //when the number is even(divisible by two), //its least significant bit is 0 }
To check for divisibility, use the modulus, or % operator. If the expression results in zero, then the first number is divisible by the second number. For example, 10%5 equals 0 because 10 is divisible by 5. When the expression does not result in 0, the number is not divisible(10%7=3, so 10 is not divisible by 7).
#! /usr/bin/env rubyn = gets.to_id = 'is not'd = 'is' if n % 7 == 0puts 'Your number ' + d + ' divisible by seven.'For more accurate answer, please define the programming/script language you plan to use..
What for? There is only a 154 inside that is divisible by 77...
All integers greater than one are divisible by prime numbers.
There does not exist a number that is divisible by all integers. The opposite is true. The number one can divided into all integers.
Write a program to find the number and sum of all integers from 100 to 300 that are divisible by 11
Because there are an infinite number of integers, there are an infinite number of integers that are divisible by 8589934581 (every 8589932481th to be exact). There is not one.
Yes. If the integers making up the number add to a number divisible by 3, then it is divisible by 3.The integers 34215 add up to 15, so it is divisible by 3.34215/3 = 11405
To find the total number of integers between 100 and 300 that are divisible by 3, we first determine the smallest and largest integers in this range that are divisible by 3. The smallest integer divisible by 3 is 102, and the largest is 297. To find the total number of integers between 102 and 297 that are divisible by 3, we calculate (297-102)/3 + 1, which equals 66. Therefore, there are 66 integers between 100 and 300 that are divisible by 3.
If you can add the integers and they add up to a number divisible by 3, then the number is divisible by 3.
All positive integers are divisible by at least one number.
Every number is divisible by any non-zero number. No, there are an infinite number of odd integers which are not evenly divisible by 7..
No. No number can be divisible by a number greater than it. 812 is greater than 6 so 6 cannot be divisible by 812.
Of the positive integers, 4 is.