Question

How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Answer 1

Let the original three-digit number be represented by \( ABC \), where \( A \), \( B \), and \( C \) are the hundreds, tens, and units digits, respectively. When the digits are reversed to form \( CBA \), the number increases by 198. We are tasked with finding how many such three-digit numbers satisfy this condition.

Step 1: Represent the Numbers

The number \( ABC \) can be written as: \[ ABC = 100A + 10B + C \] The number \( CBA \) can be written as: \[ CBA = 100C + 10B + A \] According to the given information: \[ CBA = ABC + 198 \] Substituting the expressions for \( ABC \) and \( CBA \): \[ 100C + 10B + A = 100A + 10B + C + 198 \]

Step 2: Simplify the Equation

Cancel the \( 10B \) terms from both sides: \[ 100C + A = 100A + C + 198 \] Rearranging the terms: \[ 100C - C = 100A - A + 198 \] Simplifying further: \[ 99C - 99A = 198 \] Dividing both sides by 99: \[ C - A = 2 \]

Step 3: Determine Possible Values for \( A \) and \( C \)

Since \( A \) and \( C \) are single-digit integers, the possible pairs \( (A, C) \) that satisfy \( C - A = 2 \) are: \[ (A, C) = (0, 2), (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), (7, 9) \] Since \( A \) must be at least 1 for the number to be a three-digit number, the pair \( (0, 2) \) is not valid. Thus, we have 7 valid combinations: \[ (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), (7, 9) \]

Step 4: Determine Possible Values for \( B \)

The digit \( B \) can be any value from 0 to 9 for each valid combination of \( A \) and \( C \).

Step 5: Total Number of Valid Numbers

Since there are 7 valid combinations for \( (A, C) \) and 10 possible values for \( B \), the total number of valid three-digit numbers is: \[ 7 \times 10 = 70 \]

Final Answer:

There are \( \boxed{70} \) such three-digit numbers.

Answer 2

A three-digit number increases by 198 when the digits are reversed. Let the digits be \( a \) (hundreds), \( b \) (tens), \( c \) (units).

Step 1: Express the difference

\[ (100c + 10b + a) - (100a + 10b + c) = 198 \] Simplifies to: \[ 99(c - a) = 198 \implies c - a = 2 \]

Step 2: Determine valid digit ranges

Since \( a \) is the hundreds digit, \( 1 \leq a \leq 9 \).
Since \( c - a = 2 \), \( c = a + 2 \). For \( c \) to be a digit \( 0 \leq c \leq 9 \), \( a \) must satisfy: \[ 1 \leq a \leq 7 \]

Step 3: Possible digit pairs \((a, c)\)

\[ (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), (7, 9) \]

Step 4: Count total numbers

The middle digit \( b \) can be any digit from 0 to 9 (10 possibilities).
Therefore total numbers = \( 7 \times 10 = 70 \).

Answer: \(\boxed{70}\)