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

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.

## closed as off-topic by Namaste, kimchi lover, The Phenotype, TheSimpliFire, Claude LeiboviciFeb 1 '18 at 10:54

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• Sum of all digits used must be divisible by 3. – user202729 Jan 13 '18 at 11:02
• I know that but how to calculate the number – Sufaid Saleel Jan 13 '18 at 11:03
• Don't need more answers. – Sufaid Saleel Jan 16 '18 at 15:14

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.

• Yes. The following facts are more or less implicitly used: (A) A number is divisible by $2$ if and only if its last digit is divisible by $2$. (B) A number is divisible by $3$ if and only if its digit sum is divisible by $3$. (C) A number is divisible by $6$ if and only if it is divisible by both $2$ and $3$. – Jeppe Stig Nielsen Jan 13 '18 at 14:52

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 :)

Let's find a divisability test for 6.

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

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.

$$\binom{5}{2} * 2^3 + \binom{5}{5} * 2^0 = 81$$