Question

The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is______ .

Answer 1

Let the number be represented as $N = abc$, where:

• $a$ is the hundreds digit,
• $b$ is the tens digit,
• $c$ is the ones digit.

We are given that the sum of the digits is 14:

$a + b + c = 14$

Case (i): All Distinct Digits

We have $a + b + c = 14$, where $a \geq 1$ and $b, c \in \\{0, 1, 2, \ldots, 9\\}$.

By trial, the following are the valid combinations of $a, b, c$ where the sum is 14:

$(6, 5, 3), (8, 6, 0), (9, 4, 1), (7, 6, 1), (8, 5, 1), (9, 3, 2), (7, 5, 2), (8, 4, 2), (7, 4, 3), (9, 5, 0)$

Thus, there are 8 valid cases where all digits are distinct.

Case (ii): Two Same, One Different

We now consider cases where two digits are the same and the third different. The equation becomes $2a + c = 14$, with possible solutions for $a$ and $c$:

$(3, 8), (4, 6), (5, 4), (6, 2), (7, 0)$

For each of these pairs, there are corresponding valid cases, taking into account the repetition of two digits. The number of such cases is:

$3!/2! \times 5 - 1 = 14 \text{ cases}$

Case (iii): All Digits the Same

Here, the equation $3a = 14$ must hold, but no integer value of $a$ satisfies this condition.

Hence, there are no cases where all digits are the same.

Total Number of Cases

Now, adding up the cases:

$8 \times 3! + 2 \times 4 + 14 = 48 + 22 = 70$

Thus, the total number of integers between 100 and 1000 where the sum of their digits equals 14 is:

70