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the number 11...122...25 has 1997 1's and 1998 2's. now how do i show that the number is a perfect square? i don't even know where to start. any help would be appreciated.
i guessed that much by seeing thatIt's and interesting question. I don't know how to prove it but the square root of that number is the 1998 digit integer whos first (leftmost) 1997 digits are all 3 and whos 1998th digit (rightmost) is 5.
if i understood AlphaNumeric's hint, i wouldn't have asked for more help. anyway,matt grime said:Alphanumeric's hint would appear to be the perfect answer.
shouldn't it be [tex]\underbrace{3.....3}_{k}5 = 3\underbrace{0....0}_{k} + \underbrace{3....3}_{k-1}5[/tex]?AlphaNumeric said:Use that [tex]\underbrace{3.....3}_{k}5 = 3\underbrace{0....0}_{k-2} + \underbrace{3....3}_{k-1}5[/tex]