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Math Help - Number theory

  1. #1
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    Number theory

    A number containing just two digits 0,6 (like:60,660,606.......)prove that the number will not be a perfect square.
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  2. #2
    Junior Member Sarasij's Avatar
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    Re: Number theory

    Quote Originally Posted by Swarnav View Post
    A number containing just two digits 0,6 (like:60,660,606.......)prove that the number will not be a perfect square.
    Let the number(N, say) consists of digits D0,D1,D2,...,Dn where each Di is from {0,6} and D0 must be 6 to be a (n+1) digited number. Now observe that both 0 and 6 are evenly divisible by 2.

    Consider 2 cases:
    1. Dn = 6
    2. Dn = 0

    In case 1 if the last digit of N is 6 i.e. Dn=6 then dividing it by 2 gives us a number (N/2) with last digit 3 and hence cannot be a perfect square since any further division of (N/2) will not yield any 2 from it i.e. N consists of a single 2 and hence it is a non-square number.

    In case 2 if the last digit of N is 0 i.e. Dn=0 then let the least consecutive sub-sequence of 0s from Dn backwards be upto Di where 1 <= i <= n,since D0 must be 6 to be a (n+1) digited number.

    Therefore,from D0 to Di-1 the number consists of mixes of 0s and 6s and from Di to Dn the number consists of all 0s. Pictorially the view is as follows:

    Number theory-capture.jpg

    Now separate D0 to Di-1 as a string of 0s and 6s and Di to Dn as a string of only 0s.

    In other words the number is as follows: N = D0...Di-1 X 10(n-i+1) = 6...6 X 10(n-i+1)

    Now by previous argument we know that D0...Di-1 can't be a perfect square and it also consists of a factor 6=2x3. So if the other part i.e "power of 10 is a perfect square[i.e if (n-i+1) is even]" then overall the number N can't be a perfect square.

    If (n-i+1) is odd then 10(n-i+1) can be factored as 2n-i+1 X 5n-i+1 where we see that the part 5n-i+1 still can't be a perfect square( as n-i+1 is odd and no other factors of 5 are there in the whole number ).

    So in that case also this required result is proved.
    Last edited by Sarasij; June 6th 2012 at 02:59 AM.
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    Re: Number theory

    Easy solution: Assume that the number ends in 6. We can do this because if it ends in -60 it cannot be a perfect square (all numbers in the form (10k)^2 must be multiples of 100) and if it ends in -600 (two or more zeros) we can cross off pairs of zeros, resulting in a number ending in 60 or 6.

    If the number ends in 6, it must end in either 06 or 66. Both of these numbers are 2 modulo 4, so a perfect square cannot end with these two digits.
    Last edited by richard1234; June 12th 2012 at 10:42 PM.
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