Mira's puzzle: Take any two different digits. Make two 2-digit numbers using them. Subtract the smaller number from the bigger number. Continue this process until you get a 1-digit number. What do you get in the end?
No matter which two different digits you choose, you will always get 9 in the end.
Example with digits 3 and 7:
Step 1: Form numbers: 73 and 37
Step 2: Subtract: 73 - 37 = 36
Step 3: Use digits 3 and 6: 63 - 36 = 27
Step 4: Use digits 2 and 7: 72 - 27 = 45
Step 5: Use digits 4 and 5: 54 - 45 = 9
Why it works: For digits a and b (where a > b):
- Number: 10a + b
- Reverse: 10b + a
- Difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b)
The difference is always a multiple of 9, and continuing this process eventually leads to 9.
Explanation
This puzzle demonstrates a mathematical property of two-digit numbers. When you form a two-digit number with digits a and b (where a > b), the number is 10a + b, and its reverse is 10b + a. The difference is (10a + b) - (10b + a) = 9a - 9b = 9(a - b), which is always a multiple of 9. Continuing this process with the digits of the difference will eventually yield 9.
Solution Steps
Choose two different digits, e.g., 3 and 7
Form two 2-digit numbers: 73 and 37
Subtract the smaller from the larger: 73 - 37 = 36
Use the digits of the result (3 and 6): 63 - 36 = 27
Use the digits of the result (2 and 7): 72 - 27 = 45
Use the digits of the result (4 and 5): 54 - 45 = 9
The final result is always 9