No, 8 is not a factor of 18, but 1, 2, 3, 6, and 9 are factors of 18.
No, but it is divisible by 9 and 2
The sum of the factors are divisible by 9. Ex: 738 7+3+8=18 18/9=2
To determine if 1368 is divisible by 9, you can sum its digits: 1 + 3 + 6 + 8 = 18. Since 18 is divisible by 9, it follows that 1368 is also divisible by 9. Thus, the answer is yes, 1368 is divisible by 9.
138 is not divisible by 18.
8/9 They are both divisible by 2
is 18 divisible by 9
No, but it is divisible by 9 and 2
The sum of the factors are divisible by 9. Ex: 738 7+3+8=18 18/9=2
72 is divisible by 8 and 1, 2, 3, 4, 6, 9, 12, 18, 24, 36, 72.
Because 8 is 2 cubed, you can test if a number is divisible by 8 by dividing it by 2 three times. If the number you reach is an integer, then the number is divisible by 8. For example, 72/2 = 36/2 = 18/2 = 9. Therefore, 72 is divisible by 8. For very large numbers, if the last three digits are divisible by 8, the number itself is divisible by 8.
Any number which is divisible by 9 contains digits which, when added together, are equal to 9 or a multiple of 9. For example, 63 is divisible by 9 because 6 + 3 = 9. 288 is divisible by 9 because 2 + 8 + 8 = 18 and 18/9 = 2.
74 is not divisible by 4; but704 is divisible by 4.74: 2x7+ 4 = 18; 2x1 + 8 = 10; 2x1 + 0 = 2 which is not 4 nor 8, so 74 is NOT divisible by 4704: 2x0 + 4 = 4 which is 4 or 8, so 704 IS divisible by 4
138 is not divisible by 18.
18 is not divisible by 5 and 7.
20 is not divisible by 18.
If the sum of a given number's digits is divisible by 3, the number is also divisible by 3. Thus: 1+4+5+8=18 18 is divisible by 3 (18/3=6), ergo is 1458 divisible by 3. Except for the above listed rule of thumb, a calculator comes in handy every now and then.
No. If the sum of (the hundreds digit times 4) plus (the tens digit times 2) plus (the ones digit [times 1]) is divisible by 8, then so is the original number: 2 x 4 + 1 x 2 + 8 x 1 = 18 which is not divisible by 8, so 218 is not divisible by 8.