1,215
A number is divisible by 27 if it can be expressed as 27 multiplied by an integer. This means any multiple of 27, such as 27, 54, 81, and so on, is divisible by 27. Additionally, for larger numbers, you can check divisibility by summing the digits of the number and confirming if that sum 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.
45
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.
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).
A number is divisible by 27 if it can be expressed as 27 multiplied by an integer. This means any multiple of 27, such as 27, 54, 81, and so on, is divisible by 27. Additionally, for larger numbers, you can check divisibility by summing the digits of the number and confirming if that sum 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.
45/5 = 9 45/9 = 5 Therefore, the smallest number divisible by both numbers is 45.
45
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.
135 135 / 3 = 45 135 / 5 = 27
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
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).
45% of 27 = 45% * 27 = 0.45 * 27 = 12.15
No. 135 is divisible by: 1, 3, 5, 9, 15, 27, 45, 135.
9
Subtract 8 times the last digit from remaining truncated number. Repeat the step as necessary. If the absolute of result is divisible by 27, the original number is also divisible by 27 Check for 945: 94-(8*5)=54; 5-(8*4)=-27 Since 27 is divisible by 27, the original no. 945 is also divisible. Check for 264681: 26468-(8*1)=26460; 2646-(8*8)=2582; 264-(8*6)=216 21-(8*6)=-27 Since 27 is divisible by 27, the original no. 264681 is also divisible. Check for 81: 8-(8*1)=0; Since 0 is divisible by 27, the original no. 81 is also divisible.