Learn about Divisibility by 4 and 6 This track will teach us how to check with the divisibility of a number by 4 and 6, so we need to know if a number in this case 230 is divisible by the numbers 4 and 6, so how do we do that? The first thing we need to remember is a number is divisible by another number is when I do the division there is no remainder left, which means when I divide the two numbers the end result is a whole number. So divisibility checks by 4, so I know it with 230 if it's divisible by 4, how do I check that? In order for a number to be divisible by 4, the last 2 digits should be divisible by 4 or should be 00 okay? That’s the trick that I’m going to use, so in this case the last 2 digits are 3 and 0, is 30 divisible by 4? Well if I divide 4 over 30 it gives me 7 and then I’m left with a remainder of 2, so that is not divisible by 4 and the last 2 digits are not 00 so the divisibility cast by 4 has failed. By 6, what I want to check for a divisibility test by 6 for example for a number like 230 I need to check the divisibility test by 6, well the trick to use is it has to be divisible by the factors of 6 which are 2 and 3 right? Divisibility by 2 says the last digit should be even right and divisibility by 3 says that the sum of all digits should be divisible by 3. So both of this has to be true, so let’s look at this and this example. In case of 230, the last digit is even, the last digit is 0 so this passes, so 230, last digit is even, even check is correct. Sum of all digits should be divisible by 3, sum of all digits is 5 right, 5 is not divisible by 3 so this check fails. In order for the number to be divisible by 6, it has to be divisible by 2 and by 3, it's not, so the answer is again no. So quickly recapping what we’ve learned, for a number to be divisible by 4 the last 2 digits should be divisible by 4 or the last 2 should be 0, in case of 230, the last 2 digits are 0 not divisible by 4 so that’s not. For a number to be divisible by 230 or divisible by 6, it has to be divisible by the factors of 6 which are 2 and 3, for it to be divisible by 2 the last digits should be even which is true but in order to be divisible by 3 sum of all digits should be divisible by 3 which in this case is 5 which is not true. So since both of these conditions were not met, the answer is still a no.