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

Let the natural number be represented as \( x = \overline{xyz} \), where \( x, y, z \) are the digits of the number. We are given that: \[ x + y + z = 15 \] We need to find all valid combinations of \( x, y, z \) where the sum equals 15 and \( x \) is the hundreds digit (i.e., \( 2 \leq x \leq 9 \)). 
Step 1: Case for \( x = 2 \): Here, \( y + z = 13 \), and the possible pairs for \( y \) and \( z \) are: \[ (4, 9), (5, 8), (6, 7), (7, 6), (8, 5), (9, 4) \] Thus, there are 6 possibilities. 
Step 2: Case for \( x = 3 \): Here, \( y + z = 12 \), and the possible pairs for \( y \) and \( z \) are: \[ (3, 9), (4, 8), (5, 7), (6, 6), (7, 5), (8, 4), (9, 3) \] Thus, there are 7 possibilities. 
Step 3: Case for \( x = 4 \): Here, \( y + z = 11 \), and the possible pairs for \( y \) and \( z \) are: \[ (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2) \] Thus, there are 8 possibilities. 
Step 4: Case for \( x = 5 \): Here, \( y + z = 10 \), and the possible pairs for \( y \) and \( z \) are: \[ (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1) \] Thus, there are 9 possibilities. 
Step 5: Case for \( x = 6 \): Here, \( y + z = 9 \), and the possible pairs for \( y \) and \( z \) are: \[ (0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0) \] Thus, there are 10 possibilities. 
Step 6: Case for \( x = 7 \): Here, \( y + z = 8 \), and the possible pairs for \( y \) and \( z \) are: \[ (0, 8), (1, 7), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (7, 1), (8, 0) \] Thus, there are 9 possibilities. 
Step 7: Case for \( x = 8 \): Here, \( y + z = 7 \), and the possible pairs for \( y \) and \( z \) are: \[ (0, 7), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (7, 0) \] Thus, there are 8 possibilities. 
Step 8: Case for \( x = 9 \): Here, \( y + z = 6 \), and the possible pairs for \( y \) and \( z \) are: \[ (0, 6), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 0) \] Thus, there are 7 possibilities. Now, the total number of possible values is: \[ 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64 \]