Problem 7 - Algebra for JEE ADVANCED

Question

Find the number of three-digit numbers from 100 to 999 including all numbers which have any one digit that is the average of the other two.

Step-by-Step Solution

Answer

There are 342 three-digit numbers fulfilling the condition.

Step 1: Understanding the Problem

We need to find three-digit numbers where one digit is the average of the other two digits. This means if the number is represented as \(xyz\), one of the equations holds: \(x = \frac{y+z}{2}\), \(y = \frac{x+z}{2}\), or \(z = \frac{x+y}{2}\) where \(x, y, z\) are digits from \(0\) to \(9\).

Step 2: Case Analysis for x

Consider \(x = \frac{y+z}{2}\). This implies \(2x = y + z\). Both \(y\) and \(z\) can be from \(0\) to \(9\), but \(x\) must satisfy \(100 \leq 100x + 10y + z < 1000\) and \(y + z = 2x\). For digit \(x\), possible values are from \(1\) to \(9\). For each \(x\), count \(y = z = x\). This yields \(x\) numbers for each valid \(x\).

Step 3: Case Analysis for y

Consider \(y = \frac{x+z}{2}\). Thus, \(2y = x + z\). \(y\) can be from \(0\) to \(9\). Determine pairs \((x, z)\) so that \(x + z = 2y\) holds and satisfies three-digit condition. Calculate for each valid \(y\) value.

Step 4: Case Analysis for z

Consider \(z = \frac{x+y}{2}\). Thus, \(2z = x + y\). Again, analyze the suitable digits \((x, y)\) leading to valid \(z\). Ensure any results comply with the three-digit number range. Calculate potential numbers for each valid \(z\).

Step 5: Summation of Cases

After individual calculations in each scenario, ensuring numbers belong to the three-digit scope and confirming no double-counting, sum results from all three cases.

Step 6: Verification

Count through each category's derived numbers, confirm there is no overlap or duplication among the cases, and ensure total unique digits fit criteria and constraints.

Key Concepts

Digit Averages

Understanding the concept of digit averages is crucial when tackling the problem of three-digit numbers. In this context, we're looking for numbers where one digit represents the average of the other two. Consider the number represented as \(xyz\). Here, possible formulas are:

• \(x = \frac{y+z}{2}\)
• \(y = \frac{x+z}{2}\)
• \(z = \frac{x+y}{2}\)

To understand these equations, remember that the average of two numbers is the sum of the numbers divided by two. Thus, if one digit of a three-digit number is half the sum of the other two, it serves as their average. This lays the groundwork for analyzing how these numbers can form a valid three-digit sequence, ranging from 100 to 999.

Case Analysis

In solving mathematical problems involving multiple conditions, 'case analysis' is a key technique. For this problem, we need to separately handle scenarios where each digit \(x\), \(y\), or \(z\) is the average of the other two digits. Through case analysis, we systematically examine each possibility:

• **Case 1**: \(x\) is the average, so \(2x = y+z\).
• **Case 2**: \(y\) is the average, so \(2y = x+z\).
• **Case 3**: \(z\) is the average, so \(2z = x+y\).

In each case, confirm the numbers are within the valid range of three-digit numbers and evaluate each possible combination to ensure no conditions are violated. This methodical approach helps in avoiding oversights and ensures completeness of the solution.

Valid Combinations

Finding valid combinations involves determining which sets of digits can form three-digit numbers under the average condition. Each digit from 100 to 999 can be split into hundreds, tens, and units:

• For \(x = \frac{y+z}{2}\), ensure \(y + z = 2x\) results in digits. Likewise for \(2y = x+z\) and \(2z = x+y\).
• Check if \(x\), \(y\), and \(z\) remain digits between 0 and 9.\(x\) ranges from 1 to 9 since \(x\) is the hundred's digit in a three-digit number.

As you iterate, make sure following constraints such as limit on digit sizes and ensuring numbers remain in a valid three-digit format. This careful checking prevents including combinations that do not satisfy the average requirement or number range.

Mathematical Problem Solving

Mathematical problem solving often involves applying diverse methods and strategies, aimed at simplest explanations and solutions for complex problems. This exercise illustrates several techniques including:

• Understanding of digit averages: framing the problem using averages.
• Splitting work via case analysis: tackling different scenarios methodically.
• Ensuring valid combinations: double-checking that each step adheres to number constraints.

By breaking down the problem into simpler parts, you not only find the solution more effectively but also enhance your understanding of mathematical problem solving. This deliberate, methodical process strengthens your logical reasoning abilities and boosts problem-solving efficiency in mathematics.