Not evenly. The quotient is 153/4 .
No, 49 cannot be divided by 3 evenly. 43 divided by 3 is equal to 16⅓. A general way of knowing if a number is divisible by 3 is by finding the sum of the individual numbers that is being divided. If the sum itself is divisible by 3, then the number is likewise also divisible. For example, the problem stated is 49/3. In 49, 4+9=13. Because 13 is not evenly divisible by 3, 49 also cannot be evenly divided by 3. In another example, the problem stated is 153/3. In 153, 1+5+3=9. Because 9 is divisible by 3, 153 is also likewise divisible by 3. 153/3=51.
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.
According to my calculations, the following are evenly divisible by 9: 108, 117, 126, 135, 144, 153, 162, 171, 180, 189, and 198.
1, 3, 9, 17, 51, 131, 153, 393, 1179, 2227, 6681, 20043
No, 153 is only divisible by: 1, 3, 9, 17, 51, 153.
Seven and 153.
Yes, 51 times.
Nope ! For ANY number to be exactly divisible by ten it HAS to end in zero !
Composite - it's exactly divisible by three. 1x153 and 3x51.
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 .
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.
No, 49 cannot be divided by 3 evenly. 43 divided by 3 is equal to 16⅓. A general way of knowing if a number is divisible by 3 is by finding the sum of the individual numbers that is being divided. If the sum itself is divisible by 3, then the number is likewise also divisible. For example, the problem stated is 49/3. In 49, 4+9=13. Because 13 is not evenly divisible by 3, 49 also cannot be evenly divided by 3. In another example, the problem stated is 153/3. In 153, 1+5+3=9. Because 9 is divisible by 3, 153 is also likewise divisible by 3. 153/3=51.
All the numbers are divisible by: 9