JEE Mathematics 2023 Question with Solution

Given: We need five-digit numbers formed from the digits $0, 1, 3, 5, 7,$ and $9$ without repetition.

Find: The number of such numbers that are greater than $40000$ and divisible by $5$.

A number divisible by $5$ must end in $0$ or $5$.

Case 1: Unit digit is $0$. Then the first digit must be greater than $4$, so it can be $5, 7,$ or $9$. After choosing the first digit, the middle three places can be filled in

$$4 \times 3 \times 2$$

ways. So, total numbers in this case are

$$3 \times 4 \times 3 \times 2 = 72$$

Case 2: Unit digit is $5$. Now the first digit must still be greater than $4$, but it cannot be $5$ again, so it can be only $7$ or $9$. The middle three places can again be filled in

$$4 \times 3 \times 2$$

ways. So, total numbers in this case are

$$2 \times 4 \times 3 \times 2 = 48$$

Adding both cases,

$$72 + 48 = 120$$

Therefore, the correct option is A.

Given: The number must be greater than $40000$ and divisible by $5$.

Find: The total count quickly.

Use the divisibility condition first. The last digit is forced to be either $0$ or $5$.

• If the last digit is $0$, then the first digit has $3$ choices: $5, 7, 9$.
• If the last digit is $5$, then the first digit has $2$ choices: $7, 9$.

In both cases, the remaining three positions are arranged in

$$4 \times 3 \times 2$$

ways.

So the total is

$$(3 + 2)(4 \times 3 \times 2) = 5 \times 24 = 120$$

This works because the middle three places are counted identically in both cases. Therefore, the correct option is A.