No, 153 is only divisible by: 1, 3, 9, 17, 51, 153.
1,3,9,17,51,153
Seven and 153.
Yes, 51 times.
612 is divisible by 4. To test divisibility by 4, if the last 2 digits of the number are divisible by 4, then the whole number is divisible by 4. Last 2 digits of 612 are 12, and 12 is divisible by 4 since 3x4 = 12, so 612 is divisible by 4. 612 / 4 = 153.
Nope ! For ANY number to be exactly divisible by ten it HAS to end in zero !
All are divisible by 3 if you don't mind that the answer is not a whole number. However, if you are looking for the number which is evenly divisible by 3 (resulting in a whole number) then it is 153. 153/3 = 51.
Not evenly. The quotient is 153/4 .
Composite - it's exactly divisible by three. 1x153 and 3x51.
1, 3, 9, 17, 51, 131, 153, 393, 1179, 2227, 6681, 20043
Yes, 153 is divisible by 9 because all the digits add up to 9 (i.e 1+5+3 = 9) As a rule of thumb, add the digits of a number together. If the sum is a multiple of 9, it is divisible by nine.
It's divisible by three, since the sum of its digits is divisible by three. So it's composite.
1,377 is divisible by: 1, 3, 9, 17, 27, 51, 81, 153, 459, 1377.