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

Question

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

The options are:

(A) $41$

(B) $80$

(C) $81$

(D) $161$

The last digit must be $2$. But I faced problem when calculating the number of number which are divisible by $3$. Somebody please help me.

Answer 1

As the number should be a multiple of $3$, the sum of digits must be divisible by $3$: as the digits $3$ and $9$ are themselves divisible by $3$, hence we should use either three or six $2$s.

Using six $2$s, there is only one number.

Using three $2$s, the number will be of the form: $$XXXXX2$$ where each $X$ represents a digit. We need to select an additional two $2$s, which can be placed in $\binom52 =10$ ways. The remaining three positions will then have two options each ($3$ or $9$). So, we have a total of: $$\binom52 \times 2^3 =80$$ and a grand total of $81$ ways.

Answer 2

Here are some hints:

You are right that it has to end with a $2$. Now the rest of the digits has to add up to a number that is divisible by $3$. So if you fill the rest of the slots of your $6$ digit numbers with $9$s and $3$s you are safe BUT the $2$ in the unit digit place make it so you have to hide two more $2$s in your number. So essentially what you are after is in how many ways you can put two more $2$s and fill the rest with $9$s or $3$s in the rest of the five digit places. Or of course you can fill all the six places with $2$s.

Hope this helps :)

Answer 3

Let's find a divisability test for 6.

\begin{equation} 1 = 1 \mod 6\\ 10 = 4 \mod 6\\ 100 = 4*10 = 4 \mod 6,\\ \text{and so on for higher powers of 10} \end{equation}

Thus, we find: a number X is divisable by 6 iff, cutting of the last digit, taking the sum of the other digits times 4 and adding the last digit the result is divisable by 6.

You are asked for a 6 digit number using only $2,3,9$. We are thus asked to find $a,b,c,d,e,f \in {2, 3, 9}$ such that $4 * (a + b + c + d + e) + f = 0 \mod 6$. As you noted, the last digit must be $2$, which you can conclude from the equation above quite easily by noticing that f must be even. So we conclude $4 * (a + b + c + d + e) = 4 \mod 6$ thus $a + b + c + d + e = {1,4} \mod 6$ and $a + b + c + d + e = 1 \mod 3$ follows. Since both $6$ and $9$ reduce modulo 6, either 2 or 5 higher digits must be equal to $2$, the rest can be chosen freely.

\begin{equation} \binom{5}{2} * 2^3 + \binom{5}{5} * 2^0 = 81 \end{equation}