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?
$(6^2-6)*8 = 240 < 1080$
Answer
It seems correct to me, though you may have made it more complicated than it needs to be.
The multiplication rule is quite appropriate here. We have four slots to fill; the first can be filled in five ways (since it can't be zero), the second can be filled in six ways, and the last two together can be filled in nine ways (as Arturo Magidin pointed out, 00 works as well). This gives us $5*6*9 = 270$ possibilities.
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