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Q: What is divisible 2 and 3?

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NO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 2 is divisible by 3

416 is divisible by 2 but is not divisible by 3.

123 is not divisible by 2 it is divisible by 3

Yes, it is divisible by 2 because it ends with 2, it is also divisible bye 3 because 1+3+2=6. 6 is divisible by 3, therefor 132 is divisible by 3.

1x6,743 not divisible by 2 no its not divisible by 3????

2+3+2+6=13 which is not divisible by 3 thus 2326 is not divisible by 3

yep it is because 2 is divisible by 2 and 3 is divisible by 3 and 0 is divisible by 5 have fun suckers!

No, but 432 is divisible by 2 and 3.

2 is NOT divisible by 3.

432 is divisible by 2 and 3.

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.

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

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