No,because 45 is not divisible by both 2 and 3. In order for a number to be divisible by 6 it has to be divisible by both 2 and 3.
The number 9 is the smallest, but there are an infinite number of answers to the question. 15, 21, 27, 33, 39, and 45 are some more.
45
45% of 27 = 45% * 27 = 0.45 * 27 = 12.15
A 4-digit number divisible by both 5 and 9 must be divisible by their least common multiple, which is 45. To find a 4-digit number divisible by 45, we need to find a number that ends in 0 and is divisible by 45. The smallest 4-digit number that fits these criteria is 1005 (45 x 22 = 990, and adding 15 gives us 1005).
Oh, dude, numbers divisible by both 3 and 9 are multiples of the least common multiple of 3 and 9, which is 9. So, any number that is a multiple of 9 is also divisible by 3 and 9. Like, it's just basic math, nothing to lose sleep over.
Sum the digits in blocks of three from right to left. If the result is divisible by 27, then the number is divisible by 27
No,because 45 is not divisible by both 2 and 3. In order for a number to be divisible by 6 it has to be divisible by both 2 and 3.
The number 9 is the smallest, but there are an infinite number of answers to the question. 15, 21, 27, 33, 39, and 45 are some more.
45/5 = 9 45/9 = 5 Therefore, the smallest number divisible by both numbers is 45.
135 135 / 3 = 45 135 / 5 = 27
45
45% of 27 = 45% * 27 = 0.45 * 27 = 12.15
A 4-digit number divisible by both 5 and 9 must be divisible by their least common multiple, which is 45. To find a 4-digit number divisible by 45, we need to find a number that ends in 0 and is divisible by 45. The smallest 4-digit number that fits these criteria is 1005 (45 x 22 = 990, and adding 15 gives us 1005).
9
No. 135 is divisible by: 1, 3, 5, 9, 15, 27, 45, 135.
Numbers divisible by nine are infinite. They start with these: 9, 18, 27, 36, 45, 54, 63 . . .