Problem Statement:
The sum of the digits of a two-digit number is 10. The number obtained by interchanging its digits is 1 less than twice the original number. Find the number.

Solution:

Step 1: Understand the problem
We are given a two-digit number. The sum of its digits is 10. When we interchange the digits, the resulting number is 1 less than twice the original number. We need to find the original number.

Step 2: Represent the number
Let's represent the original two-digit number as 10a + b, where 'a' represents the tens digit and 'b' represents the units digit.

Step 3: Express the given conditions
According to the problem, the sum of the digits is 10. Therefore, we have the equation:
a + b = 10

When we interchange the digits, the resulting number is 1 less than twice the original number. This can be expressed as:
10b + a = 2(10a + b) - 1

Step 4: Solve the equations
To solve the system of equations, we can either substitute the value of 'a' from the first equation into the second equation or vice versa. Let's substitute the value of 'a' from the first equation into the second equation:

10b + (10 - b) = 2(10(10 - b) + b) - 1
10b + 10 - b = 20(10 - b) + 2b - 1
9b + 10 = 200 - 20b + 2b - 1
9b + 10 = 200 - 18b - 1
9b + 10 = 199 - 18b
9b + 18b = 199 - 10
27b = 189
b = 7

Substituting the value of 'b' into the first equation:
a + 7 = 10
a = 3

Therefore, the original number is 37.

Step 5: Verify the solution
To verify our solution, let's check if the given conditions hold true:

1. The sum of the digits is 10:
3 + 7 = 10 (True)

2. When we interchange the digits, the resulting number is 1 less than twice the original number:
10b + a = 2(10a + b) - 1
10(7) + 3 = 2(10(3) + 7) - 1
70 + 3 = 2(30 + 7) - 1
73 = 2(37) - 1
73 = 74 - 1
73 = 73 (True)

Therefore, the number 37 satisfies all the given conditions.

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