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42 is divisible by 2, 3 and 6 but not by 9.

42 is divisible by 2 and 6.

42 is divisible by: 1 2 3 6 7 14 21 and 42.

42 is divisible by 2, 3, 6, & 7

Out of that list, 2, 3 and 6

No , the answer would be 42 and 2/3

Yes. (all of them)

Yes, 3 x 14 = 42. It is also divisible by these numbers: 1 2 3 6 7 21 and 42.

1, 2, 3, 6, 7, 14, 21 and 42.

1, 2, 3 and 6.

1, 2, 3, 6, 7, 14, 21, 42.

1, 2, 3, 6, 7, 14, 21, 42.

2, 3, 6, 7, 14 and 21

42 is a composite number as it is divisible by 1, 2, 3, 6, 7, 14, 21 and 42.

42, 48, 54, and 60 is. -- If it's 'even' then it must also be divisible by 2 . -- If it's divisible by 3 and also by 2, then it must be divisible by 6 . -- The multiples of 6 between 40 and 60 are 42, 48, 54, and 60 .

No. 483 is not divisible by 6.A number is divisible by 6 if it is divisible by both 2 and 3.It is divisible by 2 if it is even and it is divisible by 3 if the sum of the digits is a multiple of 3.483 is not divisible by 6 since it is not divisible by 2 although it is divisible by 3.

You are looking for the prime factorization of 42. Since 42 is definetely even, we conduct a test on the divisibility of 3 first. 4 + 2 = 6, and 6 mod 3 = 0. Therefore, it is divisible by 6. 6 factors to 3 and 2, and when 42/6 = 7, which is a prime. Therefore, the product is: 3 * 2 * 7.

It is divisible by any of its factors which are: 1, 2, 3, 6, 7, 14, 21 and 42

2, 3, and 6 . . . . . yes 4, 5, 9, and 10 . . no

To be divisible by 6, the number must be divisible by both 2 and 3:To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8};To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3.As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3.If the number is not divisible by 2 or 3 (or both) then the number is not divisible by 6.examples:126Last digit is even so it is divisible by 2 1 + 2 + 6 = 9 which is divisible by 3, so it is divisible by 3â†’ 126 is divisible by both 2 and 3, so it is divisible by 6124Last digit is even so it is divisible by 2 1 + 2 + 4 = 7 which is not divisible by 3, so it is not divisible by 3â†’ 126 is divisible by 2 but not divisible by 3, so it is not divisible by 6123Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 3 = 6 which is divisible by 3, so it is divisible by 3â†’ 123 is divisible by 3 but not divisible by 2, so it is not divisible by 6121Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 1 = 4 which is not divisible by 3, so it is not divisible by 3â†’ 121 is not divisible by either 2 or 3, so it is not divisible by 6

First, using your 2 divisibility rule, we find that both are even, therefore both are divisible by 2. Now, using the 3 divisibility rule, we know that 1 + 2 + 8 + 1 = 12, and 4 + 2 = 6, both multiples of 3. Thus, it must be divisible by 3 and 2, and it must divisible by 6! If so, then dividing both by 6 yields:1218/6 = 203, 42/6 = 7.Now both values are prime, therefore, 6 is the GCF of 1218 and 42.

yes it is divisible by 2 3 and 6

If it is divisible by 2 and 3, it is divisible by 6.

If a number is divisible by 2 and 3, it is divisible by 6.

if a number is divisible by 2 and 3 then its divisible by 6