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.
yep it is because 2 is divisible by 2 and 3 is divisible by 3 and 0 is divisible by 5 have fun suckers!
1x6,743 not divisible by 2 no its not divisible by 3????
Using the tests for divisibility:Divisible by 3:Add the digits and if the sum is divisible by 3, so is the original number: 2 + 3 + 4 = 9 which is divisible by 3, so 234 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. 234 is even so 234 is divisible by 2.Divisible by 3:Already shown above to be divisible by 3. 234 is divisible by both 2 & 3 so 234 is divisible by 6Divisible by 9:Add the digits and if the sum is divisible by 9, so is the original number: 2 + 3 + 4 = 9 which is divisible by 9, so 234 is divisible by 9Thus 234 is divisible by all 3, 6 & 9.
1, 3, 9, 379, 1137, 3411.
3411 is a composite number. A simple way to check is to sum its digits, 3+4+1+1=9, which shows that 3411 has 3 and 9 as factors so it is definitely not a prime. The prime factors of 3411 are 3x3x379.
123 is not divisible by 2 it is divisible by 3
416 is divisible by 2 but is not 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.
NO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 2 is 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!
2+3+2+6=13 which is not divisible by 3 thus 2326 is not 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 3→ 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 3→ 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 3→ 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 3→ 121 is not divisible by either 2 or 3, so it is not divisible by 6
1x6,743 not divisible by 2 no its not divisible by 3????
6 and its multiples are divisible by 2 and 3.
2 is NOT divisible by 3.