Special Square
Number Theory
Level
pending
How many natural numbers satisfy ALL of the following conditions: (a) it is a four-digit number, (b) it is a perfect square, (c) its first two digits are the same, (d) its last two digits are the same?
The answer is 1.
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1 solution
The four-digit number is of the form : aabb. When this is divided by 11, the result is : a0b. As the original number is a perfect square, this number should be divisible by 11. This is possible only if a + b = 11. Moreover, a0b/11 should be a perfect square. Its square root will be a two-digit number with first digit (a - 1) and second digit b. The sum of the two digits will then be : (a - 1) + b = 10. Note that the only two-digit perfect square with the sum of its digits 10 is 64. So the only four-digit perfect square of the form aabb is the square of 64, which is 7744. Hence, the answer is 1. ;-)