Find the number of all four-digit positive integers that are divisible by four and are formed by the digits 0,1,2,3,4,5.
The combination for all numbers would be $6^4$, but we have a few roadblocks to account for. First off 0 must be taken into account. If 0 were to be the first number it would only be a three digit number therefore:
$6^4-6^3=1080$
So we know that the number of possibilities that are divisible by 4 is less than 1080.
This is where I get stuck. We must account for the numbers that are divisible by 4. For a four digit number we have four place holders _ _ _ _. The first two placeholders do not matter. So for those locations we can denote $6^2$.
However I must account for the first placeholder. 0 cannot be a placeholder, so I'm not sure how to denote its possibility from here. I have a two element variation with repetition from {0,1,...5}. But I must account for the zero. If I simply had two variations that did not account for zero it would be $6^2$. So is it possible for me to use the same approach I used earlier?
$6^2-6^1$
The last two placeholders determine divisibility. In order for the four-digit number to be divisible by 4 the number created by the last four digits must also be divisible by 4.
From 0,1,2,3,4,5,6 we have
$4,8,12,16,20,24,28,32,36,40,44,48,52,56$
and from those selections we have $04,12,20,24,32,40,44,52$ which gives us 8 possibilities.
I'm a little confused when to use the multiplication rule so I'm not sure if this is acceptable.
If my work is right would $(6^2-6)*8$ be the correct answer?
$(6^2-6)*8 = 240 < 1080$
00
for the last two digits. $\endgroup$