The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is _____.
Show 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.
We are given a three-digit number represented as \( x = \overline{xyz} \), where \( x, y, z \) are digits. The sum of these digits must satisfy the condition:
\[
x + y + z = 15
\]
Furthermore, as \( x \) is the hundreds digit, it must be within the range \( 2 \leq x \leq 9 \).
Step 2: Enumeration of Combinations per Hundreds Digit
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: Total Count of Valid Numbers
The sum of all valid combinations is:
\[
6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64
\]
