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murshid_islam
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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 thatuart said:It'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]
A perfect square is a number that is the result of multiplying an integer by itself.
One way to prove that a number is a perfect square is by taking the square root of the number and checking if the result is an integer. If it is, then the number is a perfect square.
There are several methods for proving a number to be a perfect square, including using prime factorization, completing the square, or using the properties of perfect squares.
Yes, a number can be both a perfect square and a prime number. For example, 4 is a perfect square (2 x 2) and also a prime number.
Yes, there is a shortcut for determining if a number is a perfect square. If the last digit of the number is 0, 1, 4, 5, 6, or 9, then the number is a perfect square. If the last digit is 2, 3, 7, or 8, then it is not a perfect square.