It is: 45
A number that is divisible by both 5 and 9 must be a multiple of their least common multiple (LCM). The LCM of 5 and 9 is 45. Therefore, any multiple of 45, such as 45, 90, or 135, will be divisible by both 5 and 9.
You can continue to generate numbers that are divisible by both 5 and 9 indefinitely. There is no upper limit.
27000
A four-digit number that is divisible by both 5 and 9 must end in 0 or 5 (for divisibility by 5) and the sum of its digits must be divisible by 9 (for divisibility by 9). One example is 1080, as it ends in 0 and the sum of its digits (1 + 0 + 8 + 0 = 9) is divisible by 9. Another example is 1260, which also meets both criteria.
9990 is one such number.
The least three-digit number that is divisible by both 5 and 9 is 135.
45/5 = 9 45/9 = 5 Therefore, the smallest number divisible by both numbers is 45.
A number that is divisible by both 5 and 9 must be a multiple of their least common multiple (LCM). The LCM of 5 and 9 is 45. Therefore, any multiple of 45, such as 45, 90, or 135, will be divisible by both 5 and 9.
You can continue to generate numbers that are divisible by both 5 and 9 indefinitely. There is no upper limit.
Every number is divisible by any non-zero number. Furthermore, 405 is evenly divisible by both 3 and 9.
99999
27000
There are no numbers that satisfy this. If a number is divisible by both 2 and 5, then it must also be divisible by 10.
A four-digit number that is divisible by both 5 and 9 must end in 0 or 5 (for divisibility by 5) and the sum of its digits must be divisible by 9 (for divisibility by 9). One example is 1080, as it ends in 0 and the sum of its digits (1 + 0 + 8 + 0 = 9) is divisible by 9. Another example is 1260, which also meets both criteria.
9990 is one such number.
Your question is impossible to answer. Any number that is divisible by both 2 and 5 will also be divisible by 10. 30 and 60 are not divisible by 9.
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 ÷ 3 = 15) and 5 (45 ÷ 5 = 9).