Yes. If a number is evenly divisible by both 3 and 4, it will also be divisible by 6.
This is because the prime factor of 3 is [3] -- in other words, 3 is a Prime number -- and the prime factors of 4 are [2, 2].
Thus, any number that is divisible by 3 and 4 will have as part of its prime factors the set [2, 2, 3]. (The smallest such number is 12, which is 2 x 2 x 3.) Since the prime factors of 6 are [2, 3] -- and [2, 3] is a subset of [2, 2, 3] -- then any such number must also be divisible by 6.
Another way to look at it is to note that all numbers which are divisible by both 3 and 4 are also divisible by 12 (which is 3 x 4). Since 12 is divisible by 6, then all multiples of 12 will also be divisible by 6.
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All whole numbers are divisible by 1. Numbers are divisible by 2 if they end in 2, 4, 6, 8 or 0. Numbers are divisible by 3 if the sum of their digits is divisible by 3. Numbers are divisible by 4 if the last two digits of the number are divisible by 4. Numbers are divisible by 5 if the last digit of the number is either 5 or 0. Numbers are divisible by 6 if they are divisible by 2 and 3. Numbers are divisible by 9 if the sum of their digits is equal to 9 or a multiple of 9. Numbers are divisible by 10 if the last digit of the number is 0.
All 4 digit numbers that are divisible by 9 are also divisible by 3. The first is 1008 and the last is 9999.
All numbers of the form 6+12x are, for all integer x.
All numbers that are divisible by 8 and 12 would also be divisible by 4.
no, 18 is divisible by 9 but not by 4.