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
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.
No remainder. It has the same rule as 3 for divisibility. Add them up and if that is divisible by 27 then the number is divisible by 27.
A number is divisible by 3 if the sum of its digits is a multiple of 3. For example, 27 = 2 + 7 = 9/3 = 3. Therefore, 27 is divisible by 3.If the sum of the digits of the number in question is divisible by three, the whole number is divisible by three.
Yes it is. In order for it to be divisible by 9, the sum of the digits in the number must also be divisible by 9. In this case - the sum of the digits is 27.
Well, because it has to be a multiple of 10, you know that it has to end with a 10. The 27 is the biggest number and will be what we focus on. To get the 27 into a multiple of 10, we have to multiply it by 10.10 x 27 = 270.270 is not divisible by 12 or 8, so we continue.Keeping in mind that the number has to be a multiple of 10 and 27, we can only add intervals of 270.270+270 = 540.540 is divisible by 27, 10 and 12, but not by 8. So lets add another 270.540+270 = 810810 is only divisible by 10 and 27, much like 270, so lets add another 270.810+270 = 10801080 is divisible by 27, 10, 12 and 8 and is the lowest common multiple.
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
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.
Yes, it is 27
1,215
972 is one of the numbers that is divisible by both.
No remainder. It has the same rule as 3 for divisibility. Add them up and if that is divisible by 27 then the number is divisible by 27.
How about 27
81 is divisible by 1, 3, 9, 27 and 81
27, 21
1080