Problem 78 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The consecutive numbers of a three digit number form a G.P. If we subtract 792 from this number, we get a number consisting of the same digits written in the reverse order and if we increase the second digit of the required number by 2, the resulting number forms an A.P. The number is (A) 139 (B) 193 (C) 931 (D) None of these

Step-by-Step Solution

Answer

The number is 931 (Option C).

Step 1: Understand the Problem

We are given a three-digit number whose digits form a Geometric Progression (G.P.). When 792 is subtracted from this number, the result is a three-digit number with the digits in reverse order. Also, if we increase the second digit by 2, the digits form an Arithmetic Progression (A.P.). We need to find this number.

Step 2: Set Up Equations for G.P.

Let the three-digit number be represented as \(100a + 10b + c\), where \(a, b, c\) are the digits. In a G.P., the ratios between consecutive terms are equal, so we have \(\frac{b}{a} = \frac{c}{b}\). Simplifying, we get \(b^2 = ac\).

Step 3: Reverse the Digits

According to the problem, subtracting 792 from our original number gives us the number with digits reversed. Therefore, we have the equation:\[100a + 10b + c - 792 = 100c + 10b + a\]. Simplify to get:\[99a - 99c = 792\]and further simplify to find \(a - c = 8\).

Step 4: Check A.P. Condition

By adding 2 to the middle digit \(b\), the number formed should be in A.P. Thus, \(a, b+2, c\) must satisfy \(2(b + 2) = a + c\). Simplify to get \(2b + 4 = a + c\).

Step 5: Solve the Equations

We now have two equations: \(a - c = 8\) and \(2b + 4 = a + c\). Substitute the value from the first equation into the second:\[2b + 4 = 8 + 2c\]Simplifying gives us \(2b = 2c + 4\), hence \(b = c + 2\).

Step 6: Find Specific Digits

Combine equations \(a - c = 8\) and \(b = c + 2\) into the general structure of the digits forming a G.P. and A.P., solving for specific values within the digit range 0-9. Since \(a - c = 8\), valid digits are \(a = 9\) and \(c = 1\). Thus, \(b = c + 2 = 3\).

Step 7: Determine the Number

Substitute the digits back into the original number format: \(100a + 10b + c = 100 \times 9 + 10 \times 3 + 1 = 931\).

Step 8: Final Check

Reverse the number 931 and subtract 792 to confirm you get 139, which matches our condition: \(931 - 792 = 139\). This successfully meets the outlined conditions.

Key Concepts

Arithmetic Progression

An arithmetic progression, or A.P., is a sequence of numbers in which the difference between any two consecutive terms is constant. This is called the common difference.
To identify an A.P., remember:

• The sequence follows a predictable pattern.
• The formula for the nth term is given by: \(a_n = a_1 + (n-1)d\), where \(a_1\) is the first term, and \(d\) is the common difference.

In the context of the problem, once we adjust one of the digits of the original three-digit number by 2, the digits then create an arithmetic progression. The adjusted sequence among the digits satisfies the condition \(2(b + 2) = a + c\), ensuring all digits form an A.P. Here, increasing the middle digit by 2 changes the sequence such that the terms are evenly spaced, conforming to the rules of arithmetic progressions. This change helps solve for specific values of the digits.

Digit Reversal

Digit reversal is when the order of digits in a number is switched, typically from front to back. For a three-digit number like 931:

• The original order is 9, 3, 1.
• After reversal, the order becomes 1, 3, 9.

In the exercise, reversing digits is essential because subtracting 792 from the original number results in this reversed sequence. Understanding digit reversal helps you see how manipulating numbers can still keep them related through mathematical operations. This method is used to test if the new number fits the criteria given by the problem statement.

Three-Digit Numbers

Three-digit numbers range from 100 to 999. They consist of three digits with each digit contributing to its overall value based on its place value:
This can be broken down as:

• The hundreds place contributes \(100 \times a\).
• The tens place contributes \(10 \times b\).
• The units place contributes \(c\).

The exercise involves three-digit numbers, with specific rules applying to their formation (like being in a geometric progression) and transformations (like through digit reversal or forming an arithmetic progression by altering digits). Here, the correct number had to satisfy all conditions of digit manipulation while staying within the specified range of 100-999, resulting in a unique solution, which in this case is 931.