The two-digit number that is equal to seven times the sum of its digits is 42.
How to Arrive at the Answer
According to the provided reference, the problem states that the number is seven times the sum of its digits. Further, it mentions that the number obtained by reversing the digits is 18 less than the original number. This leads to the conclusion that the original number is 42. Let's break down the logic:
- Let the two-digit number be represented as 10*X + Y, where X is the tens digit and Y is the units digit.
- The sum of the digits is X + Y.
- The problem states that the number is equal to seven times the sum of its digits: 10X + Y = 7(X+Y).
- Also, when the digits are reversed (10Y + X), this new number is 18 less than the original number: 10Y + X = 10*X + Y - 18.
- Solving these two equations for X and Y leads to X = 4 and Y = 2.
- Thus, the original number 10X + Y becomes 10\4 + 2 = 42.
Summary
The number we are looking for is 42. It fulfills the condition of being seven times the sum of its digits (4+2=6; 6 * 7 = 42), and when the digits are reversed, the new number is 24, which is 18 less than 42.
