All multiples of 147 are divisible by 147: 147, 294, 441, 588, 735, ...
All multiples of 49 are divisible by it. Ex: 49, 98, 147 etc
Yes, it is. Any integer can be divided by any other integer.
Check the photo please
Both of these numbers are divisible by 3 and 7. The factors of 105 is 21 x 5 and that of 147 is 21 x 7. So the LCM is 21 x 35 = 735.
All multiples of 147 are divisible by 147: 147, 294, 441, 588, 735, ...
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.
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
Yes, it is. Any integer can be divided by any other integer.
1, 3, 7, 9, 21, 49, 63, 147, 441
The GCF of 105 and 147 is 21. Dividing 21 into each of these gives 5 and 7 respectively. These are not further divisible by any number except for 1.