The number is 90.
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
100 is divisible by (the integer factors of 100 are):1, 2, 4, 5, 10, 20, 25, 50, 100
There are 8 possibles.
The number is 90.
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
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.
only 50, 70, 98, and 100
87 is divisible by 1, 3, 29, and 87. 100 is divisible by 1, 2, 4, 5, 10, 20, 25, 50, and 100, therefore the only number divisible by 87 and 100 is 1.
The number 100 is divisable by 2 and 50, 4 and 25, 5 and 20 and 10
100 is divisible by (the integer factors of 100 are):1, 2, 4, 5, 10, 20, 25, 50, 100
There are 8 possibles.
The first whole number divisible by 3 is 102 and the last one 399. Let n be the number of whole numbers between 102 and 399 102 + (n - 1)x3 = 399 (this is an arithmetic progression) Solving n, n-1 = (399 - 102)/3 = 99 n = 100 Since even whole numbers among these 100 will be divisible by 6, the number not divisible is half of 100, i.e. 50. regards, lpokbeng
100 is divisible by these nine numbers: 1 2 4 5 10 20 25 50 and 100.
The number that is between 40 and 50 and is divisible by both 3 and 5 is 45. To determine if a number is divisible by both 3 and 5, you must ensure it is divisible by both 3 and 5 without leaving a remainder. In this case, 45 meets this criteria as it is divisible by both 3 (45 รท 3 = 15) and 5 (45 รท 5 = 9).
The number is 48.