There are 63 cards numbered from 1 to 63.

**Quantity I:** Probability of drawing a card such that if its digits are reversed, the new number is exactly 36 greater than the original number.

**Quantity II:** Probability of drawing a card whose number is divisible by 8 but not by 16.

Compare Quantity I and Quantity II.

Question

There are 63 cards numbered from 1 to 63.

**Quantity I:** Probability of drawing a card such that if its digits are reversed, the new number is exactly 36 greater than the original number.

**Quantity II:** Probability of drawing a card whose number is divisible by 8 but not by 16.

Compare Quantity I and Quantity II.

Accepted Answer

**Solution:**

1. Let the two-digit number be $10x + y$, where $x$ is the tens digit and $y$ is the units digit.

2. The reversed number is $10y + x$.

3. According to the problem, the reversed number is 36 more than the original number:
   $$10y + x = 10x + y + 36$$

4. Simplify the equation:
   $$10y + x - 10x - y = 36$$
   $$9y - 9x = 36$$
   $$y - x = 4$$

5. This means the units digit is 4 greater than the tens digit.

6. Now, find all such numbers between 1 and 63:
   - For $x=0$, $y=4$ → 04 (considered as 4)
   - $x=1$, $y=5$ → 15
   - $x=2$, $y=6$ → 26
   - $x=3$, $y=7$ → 37
   - $x=4$, $y=8$ → 48
   - $x=5$, $y=9$ → 59

All these numbers are less than or equal to 63.

7. Total such numbers = 6.

8. Probability for Quantity I:
   $$\frac{6}{63}$$

9. For Quantity II, numbers divisible by 8 but not by 16 between 1 and 63:
   - Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64 (64 > 63, exclude)
   - Multiples of 16: 16, 32, 48, 64 (exclude these)

Remaining numbers: 8, 24, 40, 56

10. Total such numbers = 4.

11. Probability for Quantity II:
    $$\frac{4}{63}$$

12. Comparing probabilities:
    $$\frac{6}{63} > \frac{4}{63}$$

**Answer:** Quantity I > Quantity II

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