JEE Mathematics 2025 Question with Solution

Given: We need three-digit natural numbers between $212$ and $999$ such that the sum of digits is $15$.

Find: The number of such natural numbers.

Let the three-digit number be represented as $\overline{xyz}$, where $x, y, z$ are digits. Then

$$x + y + z = 15$$

and since $x$ is the hundreds digit, it must satisfy $2 \leq x \leq 9$.

Now count possible pairs $\left(y, z\right)$ for each value of $x$:

• For $x = 2$, $y + z = 13$, giving $6$ combinations.
• For $x = 3$, $y + z = 12$, giving $7$ combinations.
• For $x = 4$, $y + z = 11$, giving $8$ combinations.
• For $x = 5$, $y + z = 10$, giving $9$ combinations.
• For $x = 6$, $y + z = 9$, giving $10$ combinations.
• For $x = 7$, $y + z = 8$, giving $9$ combinations.
• For $x = 8$, $y + z = 7$, giving $8$ combinations.
• For $x = 9$, $y + z = 6$, giving $7$ combinations.

Therefore, the total number of valid numbers is

$$6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64$$

Hence, the number of natural numbers is $64$.

Given: We are asked to count natural numbers between $212$ and $999$ whose digits add to $15$.

Find: The total number of such numbers.

Let the number be $xyz$, where $x, y, z$ are digits. Then

$$x + y + z = 15$$

with constraints

$$x \in [2,9], \quad y, z \in [0,9]$$

Set

$$x' = x - 2$$

so the equation becomes

$$x' + y + z = 13$$

where $0 \leq x' \leq 7$.

First count all non-negative integer solutions without upper bounds:

$$\text{Number of solutions} = \binom{13+3-1}{3-1} = \binom{15}{2} = 105$$

Now remove invalid cases.

If $x' \geq 8$, let $x'' = x' - 8$. Then

$$x'' + y + z = 5$$

so the number of such solutions is

$$\binom{5+2}{2} = 21$$

If $y \geq 10$, let $y' = y - 10$. Then

$$x' + y' + z = 3$$

so the number of such solutions is

$$\binom{3+2}{2} = 10$$

Similarly, if $z \geq 10$, the number of solutions is

$$10$$

There are no overlap cases, because any two such violations together make the remaining sum negative.

Therefore,

$$105 - (21 + 10 + 10) = 64$$

So, the number of natural numbers between $212$ and $999$ whose digits sum to $15$ is $64$.