Question

Consider four-digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

• A. 3
• B. 2
• C. 4
• D. 0
• E. 1

Hint:

For patterned numbers, express in algebraic form and use divisibility and perfect square conditions to limit search.

Answer 1

The Correct Option is A

Numbers are of form \(\overline{aabb} = 1100a + 11b = 11(100a + b)\). For divisibility by 11, perfect square must be \( 11 \times \text{square} \) → requires square part to have factor 11, so \( 100a + b = 11k^2 \). Testing \( a = 1\) to \(9\) with \( b = 0\) to \(9\), only three perfect squares emerge: \( 1156 = 34^2\), \( 7744 = 88^2\), \( 4489 = 67^2\).