All multiples of 147 are divisible by 147: 147, 294, 441, 588, 735, ...
1, 3, 49, and 147
All multiples of 49 are divisible by it. Ex: 49, 98, 147 etc
2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 49, 84, 98, 147, 196, 294.
If a number is divisible by 2 and 3, it is divisible by 6.
All multiples of 147 are divisible by 147: 147, 294, 441, 588, 735, ...
1, 3, 49, and 147
1, 3, 49, and 147
Yes. 1,323 / 9 = 147
All multiples of 49 are divisible by it. Ex: 49, 98, 147 etc
It i.s easy to see that 147 is composite. Simply add up the digits. 1 + 4 + 7 = 12. 12 is divisible by three, therefore 147 is divisible by three ! This trick also works for division by nine.
To determine if 441 is divisible by 3, you can add the digits of 441 together: 4 + 4 + 1 = 9. Since 9 is divisible by 3, then 441 is also divisible by 3. This is because a number is divisible by 3 if the sum of its digits is divisible by 3.
No, 147 divided by 6 is 24.5
147 has factors other than 1 and itself, so it is not a prime number. For example, 147 is divisible by 3. (You can quickly check this by adding up the digits of 147: 1 + 4 + 7 = 12. If the sum of the digits is divisible by 3, so is the number.) Therefore, 147 is a composite number.
Yes. The answer is 49.
No, they're both divisible by three
2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 49, 84, 98, 147, 196, 294.