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?
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?
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For patterned numbers, express in algebraic form and use divisibility and perfect square conditions to limit search.
Updated On: Mar 24, 2026
- 3
- 2
- 4
- 0
- 1
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The Correct Option is A
Solution and Explanation
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: \( 1156 = 34^2\), \( 7744 = 88^2\), \( 4489 = 67^2\).
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