The sum of the digits of a two-digit number is 10. If 18 is subtracted from this number, the digits of the resulting number are equal. What is the original number?

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**Solution:**

1. Let the two-digit number be represented as $10a + b$, where $a$ and $b$ are the digits.

2. According to the problem, the sum of the digits is 10:
   $$a + b = 10 	ag{1}$$

3. When 18 is subtracted from the number, the resulting number has equal digits. Let the resulting number be $11x$ (since digits are equal, it can be written as $10x + x = 11x$).

4. So,
   $$10a + b - 18 = 11x 	ag{2}$$

5. From equation (1), possible pairs $(a,b)$ with sum 10 are:
   - (1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2), (9,1)

6. Check each pair to see if subtracting 18 results in a number with equal digits:
   - For (7,3): Number = 73
     $$73 - 18 = 55$$
     Digits of 55 are equal.

7. This satisfies the condition. Hence, the number is 73.

**Answer:** $73$

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