510.
To be divisible by 3, it must be a multiple of 3.
Thus the required number is a multiple of both 3 and 5, which will be a multiple of their lcm: lcm(3, 5) = 15.
→ 495 ÷ 15 = 33 → first multiple of 15 greater than 495 is 15 x 34 = 510
→ 525 ÷ 15 = 35 → last multiple of 5 less than 525 is 15 x 34 x 510
→ number required is 510.
510
There are multiple numbers which meet this criteria. 71,73,74,75,77,78,79,81,82,83,85,86,87,89,91,93,94,95,97,98 & 99.
25 is the number not less than 20, not greater than 40, not divisible by 2 3 or 7, and not a prime number.
No number meets these criteria. The number must be between 41 and 49. The only number in that range that is evenly divisible by 11 is 44, but 44 is even, not odd.
No number is less than 20 and greater than 40, but if you switch them around, your answer is 25.
510
There are multiple numbers which meet this criteria. 71,73,74,75,77,78,79,81,82,83,85,86,87,89,91,93,94,95,97,98 & 99.
No. To be divisible by a number it must be greater than, or equal to, that number. 10 is less than 66, so 10 cannot be divisible by 66.
25 is the number not less than 20, not greater than 40, not divisible by 2 3 or 7, and not a prime number.
Every number is divisible by 4. A number less than 4 gives a quotient less than 1 when divided by 4. A number greater than 4 gives a quotient greater than 1 when divided by 4. 4 divided by 4 = 1.
No number meets these criteria. The number must be between 41 and 49. The only number in that range that is evenly divisible by 11 is 44, but 44 is even, not odd.
30
No number is less than 20 and greater than 40, but if you switch them around, your answer is 25.
Twenty-one over six.
84
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.
60