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
138 is not divisible by 18.
8/9 They are both divisible by 2
Any number whose digits add up to a multiple of three. Example: 189 is divisible by 3 because 1+8+9=18 and 18 is divisible by 3.
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
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.
20 is not divisible by 18.
18 is not divisible by 5 and 7.
138 is not divisible by 18.
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.