Question

What is the probability that a two-digit positive integer N has the property that the difference of N and the number obtained by reversing the order of its digits is a perfect cube?

• A. 4/45
• B. 5/15
• C. 6/45
• D. 7/45

Hint:

The difference between a two-digit number and its digit-reversal is always \( 9(a-b) \); find which digit gaps make this a perfect cube.

Answer 1

The Correct Option is D

Step 1: Set up N and its reverse.
Let \( N = 10a + b \), where \( a \) is the tens digit from 1 to 9 and \( b \) is the units digit from 0 to 9.
The digit-reversed number is \( 10b + a \).

Step 2: Find the difference.
\( N - (10b+a) = 9(a-b) \), always a multiple of 9.

Step 3: Find which multiples of 9 are perfect cubes.
Perfect cubes near this range are 0, 1, 8, 27 and 64.
Only 0 and 27 are multiples of 9, so \( a - b \) must be 0 or 3 in size.

Step 4: Fix the sample space.
Take the sample space as the ways to choose two distinct digits from 0 to 9 for the pair \( (a,b) \), giving \( \binom{10}{2} = 45 \) equally likely cases.

Step 5: Count the favourable cases.
Since the two digits must be distinct, only a gap of 3 qualifies: (0,3), (1,4), (2,5), (3,6), (4,7), (5,8) and (6,9), which is 7 pairs.

Final Answer:
The probability is 7/45. \[ \boxed{\dfrac{7}{45}} \]