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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Originally Posted by Swarnav

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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Let the number(N, say) consists of digits D₀,D₁,D₂,...,D_n where each D_i 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. D_n = 6
2. D_n = 0

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

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

Attachment 24031

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

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

Now by previous argument we know that D₀...D_i-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 2^n-i+1 X 5^n-i+1 where we see that the part 5^n-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.

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 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.