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Using only the digits $2$, $3$ and $9$, how many six digit numbers can be formed which are divisible by $6$?

My attempt :

To be divisible by $6$, the number has to be divisible by $2$ and $3$. Thus, the one's digit has to be $2$ only. Now we need to arrange $2$, $3$, $9$ in the left over five digits' places such that their sum is $3z-2$, where $z \in \mathbb{Z}^+$. But I am not able to proceed further. Can someone help?

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    The "divisible by 3" part limits the number of 2's you are allowed to use, but that's also the only thing that really happens. – 2017-01-14
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    Can we use only one digit at all 6 places or we have to choose all digits at least once? – 2017-01-14
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    @KanwaljitSingh you can also use one digit at all 6 places – 2017-01-14

3 Answers 3

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As you pointed it, the unit number has to be a two.

As the digits $3$ and $9$ contribute to $0$ for the sum of the digits modulo $3$, the number of two's has to be a multiple of $3$.

  • $6$ two's : only one possible number;
  • $3$ two's : you have to choose the positions of the $2$ two's other than the unit number, you have $\binom{5}{2}$ ways to do this; then you are left with $3$ places where you can put a $3$ or a $9$, so you have $2^3$ ways to do this; so the number of possibilities is $$\binom52\times2^3=80$$

So the total number of numbers you can form is $81$.

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Suppose that there are $k$ twos, $m$ threes and $n$ nines

The remaining five digits must sum to $3z-2$ as you said. Therefore $2k+3m+9n=3z-2$

Of course all the threes and nines make up part of the $3z$ in the equation. We can simplify things by letting $z=r+m+3n$. Then the equation is

$2k+3m+9n=3(r+m+3n)-2$

This simplifies to

$2k = 3r-2$

We can see that $2k+2$ must be a multiple of three. Examine all cases $1 \leq k \leq 5$, we can see that this only happens when $k=2$ and $k=5$

Now you know the limited cases for $k$ which you must consider. Can you finish this and find how many combinations of threes and nines there are for each value of $k$?

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You can use here divisibility rule of 6. For divisible by 6 we have number to be divisible by 2 and 3 both.

For divisible by 2 last digit should be 2. So only 1 way to arrange last digit.

Here some things to notice -

If number of 2's = 1 or 2 or 4 or 5 included last 2 then number formed not divisible by 6.

So we have left with only 2 cases.

Case 1- We have three 2's.

Last digit 2. Then to fill two 2's at any 2 places and at remaining 3 places we have 2 options(3,9) for each place.

$\binom{5}{2} \times 2 \times 2 \times 2 = 80$

Case 2- We have all 6-digits as 2.

Then only 1 way.

Total ways = 80 + 1 = 81