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

The correct answer is option (C): \( \tfrac{4}{45} \)

Let the two-digit integer N be represented as N = 10a + b, where a and b are digits (with 1 ≤ a ≤ 9 and 0 ≤ b ≤ 9). The number obtained by reversing the digits of N is 10b + a.

The sum of N and its reversal is
(10a + b) + (10b + a) = 11a + 11b = 11(a + b).

For this sum to be a perfect square, 11(a + b) must be a perfect square. Since 11 is prime, a + b itself must supply the factor 11 — i.e. a + b = 11·k² for some integer k. Because 1 ≤ a + b ≤ 18, the only possibility is a + b = 11 (with k = 1).

Now list all digit pairs (a,b) with a + b = 11 and 1 ≤ a ≤ 9, 0 ≤ b ≤ 9:

(2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2). There are 8 such two-digit integers.

Total two-digit integers = 90 (from 10 to 99). Therefore the required probability is
8 / 90 = 4 / 45.

Final Answer: \( \tfrac{4}{45} \)