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.
The number 105 is divisible by several integers, including 1, 3, 5, 7, 15, 21, 35, and 105 itself. To determine divisibility, you can check if the number can be divided by these integers without leaving a remainder. For example, 105 divided by 3 equals 35, which is an integer, indicating that 105 is divisible by 3.
No; since the last digit is 5, it is divisible by 5.
Yes, 105 is divisible by 3. To determine this, you can add the digits of 105 (1 + 0 + 5 = 6), and since 6 is divisible by 3, it follows that 105 is also divisible by 3. The result of dividing 105 by 3 is 35.
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.
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.