Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeating?

my explanation: total number of permutations with 0, 1, 2, 3, 4 and 5 to have 5 digits = (includes numbers with 0 at the beginning)

now let us find the numbers with zero at the beginning =

so total no of digits = 720 - 120 = 600

but this is not the answer? the correct answer is 216.

can somebody correct me whats wrong with my approach and suggest a better solution.

share|cite|improve this question
up vote 1 down vote accepted

You should note that not every five-digit number made of 0,1,2,3,4,5 is divisible by 3.

A number is divisible by 3 if the sum of its digits is. Since $0+1+2+3+4+5 = 15$, which is divisible by $3$, and you need only 5 digits, you must leave our a number that's divisible by 3, namely $0$ or $3$.

If you leave out $0$, you end up with five numbers, which can be ordered in $5! = 120$ ways.

If you leave out $3$, you end up with $120$ numbers again, but $\frac{1}{5}$ of them would begin with $0$ so they actually have $4$ digits. This means you have to rule out $\frac{120}{5} = 24$ numbers.

To sum up, you have $120 + 120 - 24 = 216$ as you mentioned.

share|cite|improve this answer

You haven’t taken into account the requirement that the numbers be divisible by $3$.

Since you’re allowed to use only the six digits $0,1,2,3,4$, and $5$ and are not allowed to repeat any digit, there are just $6$ possible sets of digits that you can use, $\{0,1,2,3,4\}$, $\{0,1,2,3,5\}$, $\{0,1,2,4,5\}$, $\{0,1,3,4,5\}$, $\{0,2,3,4,5\}$, and $\{1,2,3,4,5\}$. An integer is divisible by $3$ if and only if the sum of its digits is divisible by $3$, so only the sets $\{0,1,2,4,5\}$ and $\{1,2,3,4,5\}$ will actually give you multiples of $3$. Every permutation of each of these sets gives you a multiple of $3$.

  • How many permutations of the digits $1,2,3,4$, and $5$ are there?
  • How many permutations of the digits $0,1,2,4$, and $5$ are there that do not start with $0$?

Thus sum of those two numbers is what you’re looking for.

share|cite|improve this answer

You only substracted the numbers with 0 at the peginning however: the divisibilty rule for three says that for a number to be a multiple of 3 the sum of its digits needs to be a multiple of three. So really it does not matter what order you put the numbers in but the sum of the numbers themselves. There are in total 6 combinations of numbers you can pick. which are

$1+2+3+4+5 = 15$

$0+2+3+4+5 = 14$

$0+1+3+4+5 = 13$

$0+1+2+4+5 = 12$

$0+1+2+3+5 = 11$

$0+1+2+3+4 = 10$

Therefore only the combinations that use 1,2,3,4,5 or 0,1,2,3,4 are divisible by 3. There are 5! of the first one = 120. However we cant pick combinations which start with zero so there are 5!-4!of the second one= 96. Add them to get 216.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.