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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?
Answer & Solution
Correct Answer:
Option
D
Any four digit number in which first two digits are equal and last two digits are also equal will be in the form 11 × (11a + b) i.e. it will be the multiple of 11 like 1122, 3366, 2244, . . . .
Now, let the required number be aabb.
Since aabb is a perfect square, the only pair of a and b that satisfy the above mentioned condition is a = 7 and b = 4 Hence, 7744 is a perfect square
Now, let the required number be aabb.
Since aabb is a perfect square, the only pair of a and b that satisfy the above mentioned condition is a = 7 and b = 4 Hence, 7744 is a perfect square
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Loginthe correct form for 4 digit no. whose first 2 digits are same and last 2 digits are same should be 11*(100a+b)