JEE Mathematics 2023 Question with Solution

Given: Six-digit numbers are to be formed using the digits $4, 5, 9$ only.

Find: The total number of such numbers divisible by $6$.

A number divisible by $6$ must be divisible by both $2$ and $3$.

For divisibility by $2$, the last digit must be $4$. For divisibility by $3$, the sum of the digits must be divisible by $3$.

So we count valid arrangements of the remaining five positions.

Case 1: All digits are the same.

For $44444$, there is only $1$ number.

Case 2: Two distinct digits.

For $\left(4,5\right)$, numbers of the form $44455$:

$$\frac{5!}{3!2!}=10$$

For $\left(4,9\right)$, numbers of the form $44499$:

$$\frac{5!}{3!2!}=10$$

Case 3: Three distinct digits.

For digits $4,5,9,4,4$:

$$\frac{5!}{3!}=20$$

For digits $4,5,9,5,5$:

$$\frac{5!}{3!2!}=5$$

For digits $4,5,9,9,9$:

$$\frac{5!}{3!2!}=5$$

For digits $4,5,9,4,5$:

$$\frac{5!}{2!2!1!}=30$$

Adding all cases,

$$1+10+10+20+5+5+30=81$$

Therefore, the total number of such numbers is $81$.

Given: The number must use only the digits $4,5,9$ and be divisible by $6$.

Find: The number of valid six-digit numbers.

Since divisibility by $6$ requires divisibility by both $2$ and $3$, the last digit must be even. Among $4,5,9$, only $4$ is even, so the last digit is fixed as $4$.

Now the first five digits can be chosen from $4,5,9$ such that their sum together with the final $4$ is divisible by $3$. The solution counts the admissible multisets and then their permutations.

• All same: $44444 \Rightarrow 1$ arrangement.
• Using $4$ and $5$: $44455 \Rightarrow \frac{5!}{3!2!}=10$.
• Using $4$ and $9$: $44499 \Rightarrow \frac{5!}{3!2!}=10$.
• Using $4,5,9,4,4$: $\frac{5!}{3!}=20$.
• Using $4,5,9,5,5$: $\frac{5!}{3!2!}=5$.
• Using $4,5,9,9,9$: $\frac{5!}{3!2!}=5$.
• Using $4,5,9,4,5$: $\frac{5!}{2!2!1!}=30$.

Hence,

$$81=1+10+10+20+5+5+30$$

So the required numerical value is $81$.