Question

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

Answer 1

Correct Answer: 70

To find the number of integers between 100 and 1000 whose digits sum to 14, consider a number in this range as a three-digit number of the form \(abc\), with \(a, b, c\) as digits. The conditions are:

• \(100 \leq 100a + 10b + c < 1000\)
• \(a + b + c = 14\)
• 1 ≤ a ≤ 9 (since a three-digit number cannot start with 0)

We can thus deduce:

• \(1 ≤ a ≤ 9\)
• \(0 ≤ b, c ≤ 9\)

For each valid value of \(a\), calculate the possible combinations of \(b\) and \(c\) such that they sum with \(a\) to 14:

• For \(a = 1\), \(b + c = 13\); valid combinations: (4,9), (5,8), (6,7), (7,6), (8,5), (9,4)
• For \(a = 2\), \(b + c = 12\); valid combinations: (3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3)
• For \(a = 3\), \(b + c = 11\); valid combinations: (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2)
• For \(a = 4\), \(b + c = 10\); valid combinations: (1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2), (9,1)
• For \(a = 5\), \(b + c = 9\); valid combinations: (0,9), (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1), (9,0)
• For \(a = 6\), \(b + c = 8\); valid combinations: (0,8), (1,7), (2,6), (3,5), (4,4), (5,3), (6,2), (7,1), (8,0)
• For \(a = 7\), \(b + c = 7\); valid combinations: (0,7), (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (7,0)
• For \(a = 8\), \(b + c = 6\); valid combinations: (0,6), (1,5), (2,4), (3,3), (4,2), (5,1), (6,0)
• For \(a = 9\), \(b + c = 5\); valid combinations: (0,5), (1,4), (2,3), (3,2), (4,1), (5,0)

Adding the number of combinations for each \(a\) gives 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 + 6 = 70.

The total number is 70. Hence, the value is within the range of 70.70, confirming correctness.

Answer 2

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