They are both divisible by 1 and 3
Yes, 27 is divisible by 1, 3, 9, 27.
No. 105 is divisible by these numbers: 1, 3, 5, 7, 15, 21, 35, 105.
Since 28 is divisible by both 2 and 7, the LCM is 28.
0 is divisible evenly by any non-zero number.
The GCF of 27, 106, and 126 is 3.
They're both divisible by 1,3,7, and 21.
972 is one of the numbers that is divisible by both.
No. To be divisible by 6, the number must be divisible by both 2 and 3. To be divisible by 2 the number must be even, ie its last digit must be one of {0, 2, 4, 6, 8}; 105's last digit is 5 which is odd so 105 is not divisible by 2. To be divisible by 3, sum the digits of the number and if the result is divisible by 3, then so is the original number. For 1-5: 1 + 0 + 5 = 6 which is divisible by 3 therefore 105 is divisible by 3. Although 105 is divisible by 3 it is not divisible by 2, thus it is not divisible by 6.
No - both are divisible by 9.
No. 105 is only divisible by: 1, 3, 5, 7, 15, 21, 35, 105.
This is 1 over 9. You just have to find out what number both 3 and 27 are divisible by and then divide each number by that until you have no other number both the numerator and denominator are both divisible by (they both have to be divisible by the same number to make it equivalent).
The multiples of 105 are divisible by 105, namely:105, 210, 315, 420, 525, 630, 735, 840, 945, 1050, ...105 is divisible by its factors: 1, 3, 5, 7, 15, 35, 105
The multiples of 105 (which are infinite) are all divisible by 105, including these: 105, 210, 315, 420, 525, 630, 735, 840, 945 . . .
No, 105 is not divisible by 9 because the sum of its digits is not a multiple of 9.
No, it is not.
No. 105 is not evenly divisible by six.
27 and 54 are composite numbers because both are divisible by 3.