When a three-digit number and the sum of its digits are added, the resulting number is 387. Find all such possible three-digit numbers.

We are tasked with finding all three-digit numbers such that when the number and the sum of its digits are added, the resulting number is 387.

Let the three-digit number be represented as \( N \). We can express \( N \) as:

\[
N = 100a + 10b + c
\]

where:
- \( a \) is the hundreds digit,
- \( b \) is the tens digit,
- \( c \) is the units digit.

The sum of the digits of \( N \) is \( a + b + c \). According to the problem, the sum of the number and the sum of its digits equals 387, so we have the equation:

\[
N + (a + b + c) = 387
\]

Substitute \( N = 100a + 10b + c \) into the equation:

\[
(100a + 10b + c) + (a + b + c) = 387
\]

Simplify the equation:

\[
100a + 10b + c + a + b + c = 387
\]

Combine like terms:

\[
(100a + a) + (10b + b) + (c + c) = 387
\]

This simplifies to:

\[
101a + 11b + 2c = 387
\]

We now need to find values of \( a \), \( b \), and \( c \) (which are digits, so \( a \in \{1, 2, \dots, 9\} \) and \( b, c \in \{0, 1, \dots, 9\} \)) that satisfy this equation.

### Solving the equation:

1. **Divide the equation by 2:**
   \[
   101a + 11b + 2c = 387
   \]
   This equation can be solved by trying values of \( a \), and finding corresponding \( b \,.after

Explore how to find all three-digit numbers where the number and the sum of its digits equal 387. This problem introduces algebraic methods for solving linear equations involving digits, suitable for students in Grades 6-8.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation