Problem 35 - Elementary and Intermediate Algebra

Question

The sum of the digits in a four-digit number is \(10 .\) Twice the sum of the thousands digit and the tens digit is 1 less than the sum of the other two digits. The tens digit is twice the thousands digit. The ones digit equals the sum of the thousands digit and the hundreds digit. Find the four-digit number.

Step-by-Step Solution

Answer

The four-digit number is 1223.

Step 1: Define Variables

Let the digits of the four-digit number be represented as follows: the thousands digit by 'a', the hundreds digit by 'b', the tens digit by 'c', and the ones digit by 'd'. Hence, the number is expressed as 1000a + 100b + 10c + d.

Step 2: Set Up Equations

We are given several conditions. The sum of the digits is 10: a + b + c + d = 10 Twice the sum of the thousands digit and the tens digit is 1 less than the sum of the other two digits: 2(a + c) = b + d - 1 The tens digit is twice the thousands digit: c = 2a The ones digit equals the sum of the thousands digit and the hundreds digit: d = a + b

Step 3: Substitute 'c' in Equations

From the equation c = 2a, substitute 'c' in the other equations. a + b + 2a + d = 10, so we have: 3a + b + d = 10

Step 4: Simplify the Second Equation

Substitute 'c' and simplify the second equation: 2(a + 2a) = b + d - 1 6a = b + d - 1 Therefore, we have: b + d = 6a + 1

Step 5: Substitute 'd' in Equation

From the equation d = a + b, substitute 'd' in other equations: 3a + b + (a + b) = 10 4a + 2b = 10 Simplify to: 2a + b = 5

Step 6: Solve for 'b'

Since we have two equations from previous steps, solve for 'b': 2a + b = 5 b + (a + b) = 6a + 1 Substituting d = a + b in the second simplified equation, we get: 3b + a = 6a + 1 Rearrange to solve for 'b': 3b = 5a + 1 b = (5a + 1)/3

Step 7: Find Integer Values

For 'b' to be an integer, 5a + 1 must be divisible by 3. Solve for possible values of 'a': If a = 1, (5*1 + 1)/3 = 2 b = 2 d = a + b = 1 + 2 c = 2a = 2 Digits: a = 1, b = 2, c = 2, d = 3

Step 8: Verify Conditions

Verify all the conditions with the digits a = 1, b = 2, c = 2, d = 3. Sum: 1 + 2 + 2 + 3 = 82(a + c) = b + d - 12(1 + 2) = 2 + 3 - 1Check if those digits satisfy all conditions to confirm: Final number: 1223

Key Concepts

Variable Substitution

Variable substitution is an essential technique in solving equations. It involves replacing one variable with another expression to make solving the equation simpler.
In our problem, we are given a four-digit number and we define each digit with a variable. For example, let:

• 'a' represent the thousands digit
• 'b' represent the hundreds digit
• 'c' represent the tens digit
• 'd' represent the ones digit

From the problem, we know that the sum of the digits is 10. Therefore, the equation can be expressed as:
\(a + b + c + d = 10\).
By using variables, we can create other equations based on the conditions given in the problem. For example, the condition that twice the sum of the thousands digit and the tens digit is 1 less than the sum of the other two digits can be written as:
\(2(a + c) = b + d - 1\).
Substituting expressions for one variable in terms of others, like using \(c = 2a\), simplifies the system of equations.

Digit Sum Problem

A digit sum problem involves finding digits of a number that satisfy given conditions.
In this exercise, the conditions are:

• The sum of the digits is 10.
• Twice the sum of the thousands digit and the tens digit is 1 less than the sum of the other two digits.
• The tens digit is twice the thousands digit.
• The ones digit equals the sum of the thousands digit and the hundreds digit.

To solve it, you start by expressing each digit with a variable. For instance, let’s name the four digits: 'a', 'b', 'c', 'd'.
Next, create equations based on the problem conditions:
\[a + b + c + d = 10\]
\[2(a + c) = b + d - 1\]
\[c = 2a\]
\[d = a + b\]
Use these equations to solve for the variables and find the correct digits.

Linear Equations

Linear equations are equations of the first degree, meaning they involve the sum of products of variables and constants.
For this problem, we used several linear equations to describe relationships between the digits:

• The sum of the digits equation: \(a + b + c + d = 10\)
• The twice the sum condition: \(2(a + c) = b + d - 1\)
• The digit relationships: \(c = 2a\) and \(d = a + b\)

Each equation represents a straight-line relationship, and we solve these by substituting one variable with another.
For example, we substitute \(c\) with \(2a\) and work through the steps to simplify the equations. By solving these linear equations step-by-step, we determined the values of the digits: 'a = 1', 'b = 2', 'c = 2', 'd = 3'. This method ensures that all conditions of the problem are satisfied.