This number seems to have the factor 11
Number Theory
Level
2
Find the only four-digit integer whose first two digits are identical, last two digits are identical, and which is the square of a natural number .
The answer is 7744.
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1 solution
The number is of the form aabb, where a and b are digits. Every two digit number whise digits are equal is the result of a one digit number multiplied by 11. A number of the form aabb must be a one digit number, b, multiplied by 11 to give bb, then added to a number of the form a00 multiplied by 11 to give aa00. The sum is aabb, and dividing aabb by eleven gives a0b.
In order for aabb to be a perfect square, we must find a number of the form a0b which is 11 times a perfect square. The first number of the form a0b which is a multiple of 11 is 209, which is 19 times 11. Adding 99 (9 * 11) gives 308, or 28 * 11. Adding 396 (36 * 11) gives 704, or 64 * 11. 64 is a perfect square, so 704 works.
704*11=7744.