JEE Mathematics 2025 Question with Solution

Given: We need the number of seven-digit numbers whose digit sum is even.

Find: The value of $m+n$ if that count is $m \cdot n \cdot 10^a$.

A seven-digit number has first digit with $9$ choices and each of the remaining six digits with $10$ choices.

So, the total number of seven-digit numbers is

$$9 \times 10^6$$

For any fixed first six digits, the seventh digit is equally likely to make the total sum even or odd. If the sum of the first six digits is even, the last digit must be even. If the sum of the first six digits is odd, the last digit must be odd. Hence, exactly half of all seven-digit numbers have even digit sum.

Therefore, required count is

$$\frac{9 \times 10^6}{2} = 4.5 \times 10^6 = 9 \times 5 \times 10^5$$

Comparing with $m \cdot n \cdot 10^a$, we get $m=9$ and $n=5$.

So,

$$m+n=9+5=14$$

Therefore, the answer is $14$.

Given: Total seven-digit numbers are counted.

Find: $m+n$.

Total seven-digit numbers:

$$9 \times 10^6$$

By symmetry, half have even digit sum and half have odd digit sum.

So the number with even digit sum is

$$\frac{9 \times 10^6}{2}=4500000=9 \times 5 \times 10^5$$

Thus, $m=9$ and $n=5$, giving $m+n=14$.

The correct answer is $14$.