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