They are both divisible by 1 and 3
No. 105 is divisible by these numbers: 1, 3, 5, 7, 15, 21, 35, 105.
If a number is divisible by both three and four, it's divisible by twelve.
A number is divisible by 6 if it is divisible both by 2 and 3 1240 is even then is divisible by 2 1+2+4+0 = 7 which is not divisible by 3 then 1240 is not divisible by 3 Thus 1240 is not divisible by 6
No. 1+9+6=16, which is not divisible by 3; 1+4+6=11, which isn't divisible by 3 either. No. 1+9+6=16, which is not divisible by 3; 1+4+6=11, which isn't divisible by 3 either.
No, only 3.
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, it is not.
26.25
No. 105 is only divisible by: 1, 3, 5, 7, 15, 21, 35, 105.
They are both divisible by 1 and 3
Yes it is, and the answer to this is 35.
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
Yes. 315 ÷ 3 = 105
105 is divisible by 1, 3, 5, 7, 15, 21, 35, 105.
105
No. 105 is divisible by these numbers: 1, 3, 5, 7, 15, 21, 35, 105.