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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Write the number as \(\overline{aabb}=11(100a+b)\) and use the fact that a perfect square divisible by 11 is divisible by 121.
Updated On: Nov 4, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Write the number algebraically:
The number is \(\overline{aabb}=1100a+11b=11(100a+b)\), with \(a\in\{1,\dots,9\}\), \(b\in\{0,\dots,9\}\).
Step 2: Use divisibility by 11:
For a perfect square to be divisible by the prime 11 it must be divisible by \(11^2=121\). So 11 must divide \(100a+b\). Since \(100\equiv1\pmod{11}\), this means \(a+b\equiv0\pmod{11}\). With \(1\le a+b\le18\), the only possibility is \(a+b=11\).
Step 3: Simplify:
With \(b=11-a\): \(100a+b=99a+11=11(9a+1)\), so the number is \(121(9a+1)\). It is a perfect square exactly when \(9a+1\) is a perfect square.
Step 4: Check the allowed values of a:
Since \(b=11-a\) must be a digit, \(a\ge2\). \(9a+1\) takes the values 19, 28, 37, 46, 55, 64, 73, 82 for \(a=2,\dots,9\). Only 64 is a square, at \(a=7\), \(b=4\).
So the number is \(121\times64=7744=88^2\). It is the only such number.
Final Answer:
The correct option is (1): 1.
The number is \(\overline{aabb}=1100a+11b=11(100a+b)\), with \(a\in\{1,\dots,9\}\), \(b\in\{0,\dots,9\}\).
Step 2: Use divisibility by 11:
For a perfect square to be divisible by the prime 11 it must be divisible by \(11^2=121\). So 11 must divide \(100a+b\). Since \(100\equiv1\pmod{11}\), this means \(a+b\equiv0\pmod{11}\). With \(1\le a+b\le18\), the only possibility is \(a+b=11\).
Step 3: Simplify:
With \(b=11-a\): \(100a+b=99a+11=11(9a+1)\), so the number is \(121(9a+1)\). It is a perfect square exactly when \(9a+1\) is a perfect square.
Step 4: Check the allowed values of a:
Since \(b=11-a\) must be a digit, \(a\ge2\). \(9a+1\) takes the values 19, 28, 37, 46, 55, 64, 73, 82 for \(a=2,\dots,9\). Only 64 is a square, at \(a=7\), \(b=4\).
So the number is \(121\times64=7744=88^2\). It is the only such number.
Final Answer:
The correct option is (1): 1.
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