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 sum of the first 50 whole numbers is 1,225.
Sum of 1st n even numbers: count*average = n * (2 + 2*n)/2 = n * (n+1) Sum = 50 * (2+100)/2 = 50*51 = 2550
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.
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
only 50, 70, 98, and 100
It is of numbers evenly divisible by it like 100.
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.
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.
100, 200, 300 and so on. All numbers ending with "00".
100
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