JEE Mathematics 2023 Question with Solution

Given: We need the total number of $3$-digit numbers formed using digits $1, 2, 3, 4, 5$ with repetition, such that the number is divisible by $6$.

Find: The number of such valid $3$-digit numbers.

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

For divisibility by $2$, the unit digit must be even. From the given digits, the possible last digits are $2$ and $4$.

For the last digit $2$: The first two digits $a$ and $b$ must satisfy

$$a + b + 2 \equiv 0 \pmod{3}$$

The extracted solution states that this gives $8$ possible combinations.

For the last digit $4$: Similarly, the first two digits must satisfy

$$a + b + 4 \equiv 0 \pmod{3}$$

The extracted solution states that this again gives $8$ possible combinations.

Therefore, the total number of valid numbers is

$$8 + 8 = 16$$

So, the required answer is $16$.

The answer key shows $8$, but the solution concludes the answer is $16$. Hence, the answer is taken from the solution.

Given: Digits available are $1, 2, 3, 4, 5$ with repetition allowed.

Find: How many $3$-digit numbers are divisible by $6$.

Split the problem into two cases based on the unit digit.

1. Unit digit = $2$

• The number is already divisible by $2$.
• Now require divisibility by $3$, so the sum of all three digits must be divisible by $3$.
• Hence,

$$a + b + 2 \equiv 0 \pmod{3}$$

According to the provided solution, this gives $8$ valid combinations.

2. Unit digit = $4$

• Again the number is divisible by $2$.
• For divisibility by $3$,

$$a + b + 4 \equiv 0 \pmod{3}$$

According to the provided solution, this also gives $8$ valid combinations.

Adding both cases,

$$8 + 8 = 16$$

Therefore, the total number of such numbers is $16$.