They're both divisible by 1,3,7, and 21.
No. 105 is only divisible by: 1, 3, 5, 7, 15, 21, 35, 105.
yes, 21 times
no. if you divided it by nine you would get a number with a big decimal. But is divisible by 5. (105/5=21)
105 is divisible by 1, 3, 5, 7, 15, 21, 35, 105.
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.
No; since the last digit is 5, it is divisible by 5.
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.
17640
No. 105 is divisible by these numbers: 1, 3, 5, 7, 15, 21, 35, 105.
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
105 is divisible by 5 because it ends in a 5. 105/5 = 21, so 5*21=105. Since 21 = 3*7, and 5*3=15 and 5*7=35, 15*7 = 105 and 35*3 = 105 too. Also, 105 * 1 = 105. 210 * 1/2 = 105, 315 * 1/3 = 105, etc. Here I multiplied 105 by 2 or 3 and then multiplied back by 1/2 or 1/3.
The multiples of 105 (which are infinite) are all divisible by 105, including these: 105, 210, 315, 420, 525, 630, 735, 840, 945 . . .