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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?

$(6^2-6)*8 = 240 < 1080$

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2  
You forgot 00 for the last two digits. –  Arturo Magidin Nov 28 '11 at 0:43

3 Answers 3

up vote 4 down vote accepted

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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@ArturoMagidin many thanks, corrected. –  smackcrane Nov 28 '11 at 0:48
  1. It is easier to count the total directly: you have five possibilities for the first digit (any digit except $0$), and six each for the remaining three digits, so the total number is $5\times 6^3$. This agrees with your count ($6^4-6^3 = 6^3(6-1) = 6^3\times 5$), but I think it's easier to just count them directly.

  2. You can use the same method for counting the possibilities of the first two digits: five possibilities for the first digit, six for the second, for a total of $5\times 6$; this is the same as your count, $6^2-6^1 = 6(6-1)$.

  3. The rest is almost correct; again, you can be a bit briefer with the count of the last two digits: the first digit can be any of the six possibilities. If the first digit is 0, 2, or 4, then the second digit must be 0 or 4 to get a multiple of $4$; so for each of the three possibilities you have two $2$-digit combinations, giving $3\times 2=6$ possibilities. If the first digit is $1$, $3$, or $5$, then the second digit must be $2$, so you have only three more. The total here is the sum of the two, so we have $6+3=9$ possibilities.

    What you forgot is that $00$ also works.

Since the choices for the first two digits are independent of the choices for the last two, so you multiply the totals. The total will then be $$5\times 6 \times 9 = 270.$$

The fact that the total number (1080) is larger than the count of those that are divisible by 4 should not be a surprise: there are lots of numbers among the 1080 4-digit numbers in which every digit is one of 0, 1, 2, 3, 4, and 5 that are not divisible by $4$. So I'm not sure what your last line is meant to represent.

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I guess the last line is kind of a sanity check. –  user17366 Nov 28 '11 at 2:45

well the fact that 44 is not considered where as it should be makes it 5*6*10=300 combinations :)

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