492 is an even number so it must be divisible by 2. Its digits total a multiple of 3 so it must be divisible by 3. Any even number that is divisible by 3 is divisible by 6. Of course, you can always divide 492 by 6. If it comes out even, it's a factor. (It does.)
No, because 6 is not a factor of 50.
6 is a factor of 246 because 6 divides evenly into 246 with no remainder.
Yes, 6 is a factor of 492. If you divide 492 by 6 you get 82 with no remainder. The fact that there was no remainder means that the first number is a factor of the second number. Taking this to the obvious conclusion; the prime factors of 492 are 2, 2, 3, and 41. 6 is 2 times 3, so it is a non prime factor, but it is a factor nontheless.
Yes. If 492 is even, it has 2 as a factor. If its digits add up to a multiple of 3, it has 3 as a factor. If it has 2 and 3 as a factor, it has 6 as a factor.
I know the answer is x(x^2+3x+6) + 4(x^2+3x+6) which factors to (x+4)(x^2+3x+6) but I don't know the rule behind this. Could someone explain it to me?
no it its not im only 11 and i know it
First you add up all of the digit's, and then divide. If 6 fits in it, then 3 also fits in it, but not 9. (For ex. Is 6 a factor of 674. 6+7+4=17, 17/6=2.8. so no, 6 is not a factor of 674) If your number is even and divisible by three, it's divisible by 6.
6 is a factor of 54.
No, 18 is a multiple of 6. 6 is a factor of 18.
n = 6, 12, 18, 24
explain how you know that an estimated quotient of 6 is too large for 416/73
no 6 is not a factor of 56