JEE Mathematics 2026 Question with Solution

Given: The number must be greater than $5000$, less than $9000$, formed using digits $\{0,1,2,5,9\}$ with repetition allowed, and divisible by $3$.

Find: The total number of such numbers.

Step 1: Thousand's place condition. The number lies between $5000$ and $9000$, so the thousand's digit can be

$$5$$

Step 2: Divisibility by $3$. For divisibility by $3$, sum of digits must be divisible by $3$. Digits available are

$$\{0,1,2,5,9\}$$

Possible remainders modulo $3$ are

$$0:\{0,9\},\; 1:\{1\},\; 2:\{2,5\}$$

Step 3: Counting valid combinations. Total valid combinations for the remaining three places satisfying divisibility is $78$.

Therefore, the required number is $78$.

Given: Repetition of digits is allowed and divisibility by $3$ depends on the sum of digits modulo $3$.

Find: How many valid numbers satisfy all conditions.

Use the hint: always work using digit sum modulo $3$. Since the thousand's digit is fixed as $5$, its remainder modulo $3$ is

$$5 \equiv 2 \pmod{3}$$

So the sum of the remaining three digits must contribute remainder $1$ modulo $3$. The available digits are grouped as follows:

$$0:\{0,9\},\; 1:\{1\},\; 2:\{2,5\}$$

The extracted solution states that the total valid combinations for the remaining three places satisfying this condition is $78$.

Hence, the total number of required numbers is $78$.