Top Answer

There are eight pairs of numbers divisible by three that sum to 150, and 75 is also divisible by three, so:

sum = 17 * 75 = 750 + 525 = 1275

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there are 17 divisible by 3 between 50 to 100 , 51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99

Solution 1public class Divisible {int num;int sum;public void calculate() {sum=0;System.out.println("Numbers divisible by 9:");for(num=50;numif(num%9==0) {System.out.println(num);sum=sum+num;}}System.out.println("Sum: "+sum);}public static void main(String a[]) {Divisible obj=new Divisible();obj.calculate();}}Solution 2public static void printNums() {// We know that 54 is the first number in the range divisible by 9.// So if we start there and add 9, we will quickly find all numbers// in the given range which are evenly divisible by 9.for( int i = 54; i System.out.println(i);}}

100 is divisible by 5 and 25, 2 and 50, 5 and 20 and 10

only 50, 70, 98, and 100

It is of numbers evenly divisible by it like 100.

100 is divisible by these nine numbers: 1 2 4 5 10 20 25 50 and 100.

There are (500-100)/2 = 200 numbers divisible by 2 between 100 and 500 counting 100 but not 500. Of these (500-100)/8 = 50 are divisible by 8. So there are 150 numbers between 100 and 500 divisible by two but not by 8. By relative primeness exactly 50 out of these 150 are divisible by 3 and therefore these 50 are exactly the ones divisible by 6 but not by 8.

The prime number that is closest to 50 and smaller than it is 47 (check it, 48 and 49 are divisible by other numbers than 1 and themselves). The prime number that is closest to 50 and is bigger than it is 53 (51 and 52 are divisible by other numbers than 1 and themselves). Therefore: 47+53=100 100 is the answer.

100 is divisible by (the integer factors of 100 are):1, 2, 4, 5, 10, 20, 25, 50, 100

No. 100 is only divisible by 5 and these other numbers: 1, 2, 4, 10, 20, 25, 50, 100.

50

over 50

20 +30 + 50

50 is evenly divisible by these numbers: 1, 2, 5, 10, 25 and 50.

It is evenly divisible by 1, 2, 4, 5, 10, 20, 25, 50 and 100.

Sum of first n natural numbers is (n) x (n + 1)/2 Here we have the sum = 100 x (101)/2 = 50 x 101 = 5050

The factors of 100 are: 1, 2, 4, 5,10, 20,25, 50, 100But all of these are not divisible in 120.The following are common to both: 1,2,4,5,10,20.

Exactly 50

The sum of the first 50 natural numbers is 1,251.

The sum of the first 50 whole numbers is 1,225.

The sum of the first 50 odd numbers is 2,500.

The sum of the first 50 even numbers is 2,550.

50 is divisible by 1, 2, 5, 10, 25 and 50 in the set of whole numbers. In the set of real numbers, 50 is divisible by any number and the answer will be a whole number only if it is divided by the 6 numbers mentioned above.

Sum of 1st n even numbers: count*average = n * (2 + 2*n)/2 = n * (n+1) Sum = 50 * (2+100)/2 = 50*51 = 2550

No. 50 is divisible by these numbers: 1, 2, 5, 10, 25, 50.