Any element of the union of the following two sets:{the set of numbers of the form 689*k, where k is an integer} and{the set of numbers of the form 10*k, where k is an integer}is evenly divisible by 689 or by 10.
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Yes.
It is divisible by 2, 5 and 10 but not the rest.
If a number is divisible by both three and four, it's divisible by twelve.
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.
Using the tests for divisibility:Divisible by 3:Add the digits and if the sum is divisible by 3, so is the original number: 6 + 8 + 4 = 18 which is divisible by 3, so 684 is divisible by 3Divisible by 6:Number is divisible by 2 and 3: Divisible by 2:If the number is even (last digit divisible by 2), then the whole number is divisible by 2. 684 is even so 684 is divisible by 2.Divisible by 3:Already shown above to be divisible by 3. 684 is divisible by both 2 & 3 so 684 is divisible by 6Divisible by 9:Add the digits and if the sum is divisible by 9, so is the original number: 6 + 8 + 4 = 18 which is divisible by 9, so 684 is divisible by 9Thus 684 is divisible by all 3, 6 & 9.