The correct answer is option (D):
5
Let's break down this problem step by step to find the correct answer.
We are given that 'ab' is a two-digit number and 'ccb' is a three-digit number, where a, b, and c are distinct digits (meaning they are all different). We also know that (ab)^2 = ccb and ccb < 300.
First, let's represent the numbers algebraically:
• ab = 10a + b
• ccb = 100c + 10c + b = 110c + b
The core equation is (10a + b)^2 = 110c + b.
Since ccb < 300, the digit 'c' must be either 1 or 2 (because 300 is the upper limit, so the hundreds digit can only be 1 or 2).
Now, let's consider the possible values of 'c':
Case 1: c = 1
If c = 1, then the three-digit number ccb has the form 11b. We need to find a perfect square (ab)^2 that equals 11b. Because ccb is a perfect square, it must also end in the same digit as its square root, which is 'b'. This means b^2 must have b as the units digit. The only digits that satisfy this rule are 0, 1, 5, and 6 (because 0² = 0, 1² = 1, 5² = 25, 6² = 36).
• If b = 0, then 110 is not a perfect square.
• If b = 1, then 111 is not a perfect square.
• If b = 5, then 115 is not a perfect square.
• If b = 6, then 116 is not a perfect square.
However, none of those perfect squares actually produce a "11b" format because the value of ab is not an integer. So, c cannot be 1.
Case 2: c = 2
If c = 2, then the three-digit number ccb has the form 22b. The possible squares that end in the value of b are 0, 1, 5, and 6.
• If b = 0, then 220 is not a perfect square.
• If b = 1, then 221 is not a perfect square.
• If b = 5, then 225 is a perfect square. The square root of 225 is 15. So, ab = 15. Since a = 1, b = 5, and c = 2, and the digits are distinct, this is a possible solution.
• If b = 6, then 226 is not a perfect square.
Since 15² = 225 and the condition that a, b, and c are distinct digits is satisfied (a = 1, b = 5, c = 2), we've found our solution.
Therefore, the value of b is 5.