Let's solve the problem step-by-step.

1. Define the Number: Let the number be $10a + b$, where:

   • $a$ is the tens digit
   • $b$ is the units digit

   We know that $10a + b < 100$, so $a$ and $b$ are digits between 0 and 9.

2. Condition 1: The Tens Digit is 2 More than the Units Digit This means: $a = b + 2$

3. Condition 2: The Remainder when Digits Reversed is Subtracted

   • If we reverse the digits, the number becomes $10b + a$.
   • The difference between the original number and the reversed number is: $(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$

4. Condition 3: The Remainder is 3 Times the Sum of the Digits

   • The sum of the digits is $a + b$.
   • Therefore, we have: $9(a - b) = 3(a + b)$

5. Set Up the Equations: Now we have the system of equations: $a = b + 2$ $9(a - b) = 3(a + b)$

6. Solve the System of Equations: Substitute $a = b + 2$ into the second equation: $9((b + 2) - b) = 3((b + 2) + b)$ Simplify each part: $9 \cdot 2 = 3(2b + 2)$ $18 = 6b + 6$ Subtract 6 from both sides: $12 = 6b$ Divide by 6: $b = 2$ Substitute $b = 2$ back into $a = b + 2$: $a = 2 + 2 = 4$

7. Find the Number: The number is $10a + b = 10 \cdot 4 + 2 = 42$.

8. Verification:

   • The number is 42, with digits 4 and 2.
   • The tens digit (4) is 2 more than the units digit (2).
   • Reversing the digits gives 24.
   • Subtracting 24 from 42 gives 18.
   • The sum of the digits is $4 + 2 = 6$, and three times this sum is $3 \times 6 = 18$, which matches our difference.

Thus, the number is 42.

Would you like additional details or have any questions?

Related Problems

1. What if the tens digit is 3 more than the units digit?
2. What if the difference is 4 times the sum of the digits?
3. How does the solution change if the number is greater than 100?
4. What if the difference is twice the sum of the digits?
5. How would the approach differ if the digits' positions didn't affect the result?

Tip: When working with digit problems, represent the number in terms of its positional value to set up equations easily.