Extending a number to a square
Given any finite sequence of digits, is it true that there's always a perfect square which begins with that sequence?
For example: this is true for the sequence because .
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1 solution
Given a sequence S, let's find a perfect square that begins with that sequence.
Suppose S = 6 , 9 , 1 , 1
Pick any number between 6 9 1 1 and 6 9 1 2 with a finite number of number after the decimal points. (Clearly, it must exist. In this case, we can choose 83.135) Squaring this number will lead to a number that starts with 6911. However, it's not an integer. To fix this, multiply by 10 until it is. Now, the resultant number's square will begin with the sequence.
83135^2 = [6911]428225.
This method will work with any sequence, so the statement is true in general.