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806 would have to be divisible by 3 and 2 (6 =3*2). OK, let us see. The last digit is even, so sure, 806 is divisible by 2. But adding the three digits together, I get 14 (=8+0+6), which is not divisible by 3. So, no, 806 is not divisible by 6. I have described the fun way. The more direct way would be doing factorization (806 = 403*2 = ? Uh oh, 403 seems to be a prime factor, because it is indivisible by 2, 3, 5, 7, 11, and ...; no, wait, it is divisible by 13. So I have 806 = 13 * 31 * 2.) or using a calculator.
The rules that I can remember are as follows.
divisible by 2, if the number is even;
3, if the sum of the digits is divisible by 3;
5, if the last digit is 0 or 5;
11, if the sum of odd digits equals the sum of even digits (e.g. 121 is divisible by 11 because 1+1 = 2);
the rule for 7 is a little difficult to remember, so I consult .mathsisfun.com/divisibility-rules.html.
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What is 756 divide by what?
756 is divisible by these numbers: 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 27, 28, 36, 42, 54, 63, 84, 108, 126, 189, 252, 378, 756.
Is 90 divisible by 2 3 4 5 6 9 and 10?
no not by 10 as 216 ends in 6 not 0 same with 5 as does not end in 5 or 0 216 = 2 x 108 = 3 x 72 = 4 x 54 = 6 x 36 = 9 x 24 --------------------------------------- Last digit of 216 is even (one of {0, 2, 4, 6, 8}) so 216 is divisible by 2 6 + 1 + 2 = 9 which is divisible by 3 so 216 is divisible by 3 Last 2 digits are 16 which are divisible by 4, so 216 is divisible by 4 Last digit is not 0 nor 5, so 216 is NOT divisible by 5 216 is divisible by both 2 and 3 so 216 is divisible by 6 6 + 1 + 2 = 9, so 216 is divisible by 9 Last digit is not 0, so 216 is NOT divisible by 10 216 is divisible by 2, 3, 4, 6, 9. 216 is NOT divisible by 5 nor 10.
Is 3 divisible by 870?
No. But 870 is divisible by 3 (because 8+7+0 = 15 -> 1 + 5 = 6 and 6 is divisible by 3). 870/3 = 290.
Is there a number divisible by 6. But not 3?
yeah3 9 15 21 27 and so on(every other multiple of three)But if you have a large number, like 755253, and don't have a calculator handy, then use sum of digits to determine if divisible by 3.Then if the number is divisible by 3 and divisible by 2, then it is also divisible by 6.So in this case: 7 + 5 + 5 + 2 + 5 + 3 = 27, which is divisible by 3, but the original number is odd, so the number is not divisible by 6.
Is a number divisible by 20 a even number?
It is divisible by 2, 5 and 10 but not the rest.
