Question

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

• A. \(\frac{2}{45}\)
• B. \(\frac{1}{15}\)
• C. \(\frac{4}{45}\)
• D. \(\frac{5}{90}\)
• E. \(\frac{5}{45}\)

Answer 1

The Correct Option is C

To solve this problem, we need to analyze the given condition: a two-digit integer \( N \) and the number obtained by reversing the order of its digits sums up to a perfect square.

Let's break this down step-by-step: 

1. Consider a two-digit number \( N \). Let the tens digit be \( a \) and the units digit be \( b \). Thus, \( N \) can be expressed as \( 10a + b \).
2. The number obtained by reversing the digits will then be \( 10b + a \).
3. The sum of these two numbers is:
   • \((10a + b) + (10b + a) = 11a + 11b = 11(a + b)\).
4. We need this sum \( 11(a + b) \) to be a perfect square.
5. The range of \( a \) is from 1 to 9 (since \( N \) is two-digit), and \( b \) ranges from 0 to 9.
6. The expression \( a + b \) will thus range from 1 (i.e., the smallest possible sum \(1+0\)) to 18 (i.e., the largest possible sum \(9+9\)).
7. Now, we need \( 11(a + b) = k^2 \), where \( k \) is an integer. Therefore, the values of \( a + b \) should be such that \( 11(a + b) \) is a perfect square.
8. We consider potential values of \( k^2 \) which are perfect squares and check if they are multiples of 11. Perfect squares less than the maximum possible sum of 198 are considered: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196.
9. Among these, the perfect squares which are multiples of 11 are 121 and 44 (since \(11 \times 11 = 121\) and \(11 \times 4 = 44\)).
10. First, for \( 11(a+b) = 121 \), we have \( a + b = 11 \).
11. Second, for \( 11(a+b) = 44 \), we have \( a + b = 4 \).
12. Now, calculate the number of possibilities for each case:
    • When \( a + b = 11 \), solutions are: (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2). That is, there are 8 possibilities.
    • When \( a + b = 4 \), solutions are: (1, 3), (2, 2), (3, 1), (4, 0). That is, there are 4 possibilities.
13. In total, there are \( 8 + 4 = 12 \) numbers that satisfy the condition.
14. The total number of two-digit numbers is 90 (from 10 to 99 inclusive).
15. Hence, the probability is \( \frac{12}{90} = \frac{4}{30} = \frac{2}{15} \).

Therefore, the probability that a two-digit positive integer \( N \) has the property that the sum of \( N \) and the number obtained by reversing the order of its digits is a perfect square is \(\frac{4}{45}\).