Question

The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is _____.

Hint:

When calculating the sum of digits, break the problem into cases based on the hundreds digit and check for the possible pairs of tens and units digits.

Answer 1

Correct Answer: 64

Step 1: Understand the Problem

We are given a three-digit number \(x = \overline{xyz} \), where \( x, y, z \) are digits. The condition is: \[ x + y + z = 15 \] Additionally, since \( x \) is the hundreds digit, \( x \) must satisfy \( 2 \leq x \leq 9 \).

Step 2: Identify Possible Combinations for Each \( x \)

• For \( x = 2 \), \( y + z = 13 \) → 6 combinations: (4, 9), (5, 8), (6, 7), (7, 6), (8, 5), (9, 4)
• For \( x = 3 \), \( y + z = 12 \) → 7 combinations: (3, 9), (4, 8), (5, 7), (6, 6), (7, 5), (8, 4), (9, 3)
• For \( x = 4 \), \( y + z = 11 \) → 8 combinations: (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)
• For \( x = 5 \), \( y + z = 10 \) → 9 combinations: (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)
• For \( x = 6 \), \( y + z = 9 \) → 10 combinations: (0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0)
• For \( x = 7 \), \( y + z = 8 \) → 9 combinations: (0, 8), (1, 7), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (7, 1), (8, 0)
• For \( x = 8 \), \( y + z = 7 \) → 8 combinations: (0, 7), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (7, 0)
• For \( x = 9 \), \( y + z = 6 \) → 7 combinations: (0, 6), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 0)

Step 3: Calculate the Total Number of Possible Values

The total number of valid combinations is: \[ 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64 \]

Final Answer: 64

Answer 2

Step 1: Understand the problem.
We are asked to find the number of natural numbers between 212 and 999 whose digits add up to 15.

Step 2: Represent the number.
Let the 3-digit number be represented as \( xyz \), where \( x, y, z \) are its digits.
For all 3-digit numbers, \( x \) ranges from 2 to 9 (since the smallest number is 212).

The condition is: \[ x + y + z = 15 \] and since \( 212 \leq xyz \leq 999 \), \( x \geq 2 \).

Step 3: Solve using combinatorics.
We will count the number of non-negative integer solutions to: \[ x + y + z = 15 \] with constraints: \[ x \in [2, 9], \quad y, z \in [0, 9] \] Let \( x' = x - 2 \). Then: \[ x' + y + z = 13, \quad \text{where } 0 \leq x' \leq 7 \]

Now we first find all non-negative integer solutions without upper bounds using the formula: \[ \text{Number of solutions} = \binom{13 + 3 - 1}{3 - 1} = \binom{15}{2} = 105 \]

Now we must subtract cases where \( x' > 7 \), \( y > 9 \), or \( z > 9 \).

Case 1: \( x' \geq 8 \)
Let \( x'' = x' - 8 \), then: \[ x'' + y + z = 5 \] \[ \text{Solutions} = \binom{5 + 2}{2} = 21 \]

Case 2: \( y \geq 10 \)
Let \( y' = y - 10 \), then: \[ x' + y' + z = 3 \] \[ \text{Solutions} = \binom{3 + 2}{2} = 10 \]

Case 3: \( z \geq 10 \)
Similarly, number of solutions = 10.

Case 4: Overlaps (inclusion-exclusion).
- \( x' \geq 8, y \geq 10 \): \( x'' + y' + z = -5 \) → No solution.
- \( x' \geq 8, z \geq 10 \): \( x'' + y + z' = -5 \) → No solution.
- \( y \geq 10, z \geq 10 \): \( x' + y' + z' = -7 \) → No solution.

Hence, total valid solutions: \[ 105 - (21 + 10 + 10) = 64 \]

Step 4: Final Answer.
The number of natural numbers between 212 and 999 whose digits sum to 15 is: \[ \boxed{64} \]