14.
Assuming dealing with only counting numbers (ie integers greater than 0):
Numbers divisible by 5 or 7 are their multiples.
50 ÷ 5 = 10 → last multiple of 5 less than 50 is 9 x 5
→ 9 numbers less than 50 are divisible by 5
50 ÷ 7 = 71/7 → last multiple of 7 less than 50 is 7 x 7
→ 7 numbers less than 50 are divisible by 7
Numbers divisible by both are those which are multiples of their lowest common multiple = 35
50 ÷ 35 = 115/35 → last multiple of 35 less than 50 is 1 x 35
→ 1 number less than 50 is divisible by both 5 and 7 and needs to be removed from both the above counts.
→ (9 - 1) + (7 - 1) = 14 numbers less than 50 are divisible by 5 or 7 but not both.
If there is no restriction on numbers being greater than 0, there are infinitely many numbers as it includes the infinite number of negative numbers which are all less than 50 and provide an infinite number of numbers divisible by 5 or 7 but not both.
How many numbers less than 700 are divisible by both 15 and 21?
19
In order to be divisible by both 2 and 3, a number must be divisible by 6.The largest multiple of 6 less than 100 is 96 ... the 16th multiple of 6.So there are 16 of them.
49
Thirty-three
12
How many numbers less than 700 are divisible by both 15 and 21?
19
19
Here is a list of whole numbers less than 100 which are evenly divisible by 7714212835424956637077849198
18 numbers are there
In order to be divisible by both 2 and 3, a number must be divisible by 6.The largest multiple of 6 less than 100 is 96 ... the 16th multiple of 6.So there are 16 of them.
Four of them.
4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96,100 ............ on and on and on to infinity and beyond LOL !!!!!!!!!
The numbers that are divisible by both 3 and 5 have to be a factor of 15. That leaves 15, 30, 45, 60, 75, and 90. There are 6 numbers less than 100 divisible by 3 and 5
49
49