Q: What is a number that is divisible by 2 and 3?

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It is divisible by 6

102 is divisible by: 1 2 3 6 17 34 51 102.

For starters, the number 3 is divisible by 3 but not by 2. Next, 9, 27 etc. All the odd multiples of 3 are divisible by 3 but not 2. So there is an infinite number of numbers that are divisible by 3 and not 2.

If a number is even (divisible by 2) and divisible by 3, then it must also be divisible by 6.

Rule: If it is divisible by 2 and by 3Numbers are divisible by 6 if they are divisible by 3 and 2. Therefore, for example, since 18 is divisible by 3 and 2, it is also divisible by 6.The divisibility rule for 6 is, it can go into the number if 2 and 3 can also.

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Since 5232 is divisible by both 2 and 3, it is divisible by 6.A number must be divisible by both 2 and 3 to be divisible by 6.The number 5232 is even, so it is divisible by 2.If you add the individual digits in the number (5+2+3+2=12) you get a number that is divisible by 3, meaning the original number (5232) is also divisible by 3.

It is divisible by 6

Any number that is divisible by both 2 and 3 is divisible by 6.

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.

To determine if a number is divisible by 6, it must be divisible by both 2 and 3. To determine if a number is divisible by 2, it should be even - in other words, it should end with 0, 2, 4, 6, or 8. To determine if a number is divisible by 3, the sum of its digits should be divisible by 3. 54,132 is an even number, so it is divisible by 2. 5 + 4 + 1 + 3 + 2 = 15, which is divisible by 3, so 54,132 is divisible by 3. Since 54,132 is divisible by both 2 and 3, it is divisible by 6.

102 is divisible by: 1 2 3 6 17 34 51 102.

For starters, the number 3 is divisible by 3 but not by 2. Next, 9, 27 etc. All the odd multiples of 3 are divisible by 3 but not 2. So there is an infinite number of numbers that are divisible by 3 and not 2.

If a number is even (divisible by 2) and divisible by 3, then it must also be divisible by 6.

Rule: If it is divisible by 2 and by 3Numbers are divisible by 6 if they are divisible by 3 and 2. Therefore, for example, since 18 is divisible by 3 and 2, it is also divisible by 6.The divisibility rule for 6 is, it can go into the number if 2 and 3 can also.

The smallest number divisible by 2 3 5 is 30.

3411 is an odd number and is not divisible by 2. but, it is divisible by 3. 3411/3 = 1137

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