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600 sixes (Posted on 2019-02-11) Difficulty: 1 of 5
Can a number consisting of 600 sixes and some zeros be a square?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Puzzle Solution Comment 3 of 3 |
We will state and prove two lammas.

LEMMA 1: 10^(2m+1) * n is not a perfect square where n is an integer and n does NOT end in a zero (or, zeros) 
PROOF:
If 10^(2m+1) * n is a perfect square, then n must possess the form 10 * t^2., so that n must end in at least one zero. 
This is a contradiction.
Consequently, any positive integer ending in an ODD number of consecutive zeroes can NEVER be a perfect square.

LEMMA 2: If the units digit of a perfect square is 6, then the tens digit must be odd.
PROOF:
If a perfect square ends in a 6, its square root must have one of the  the forms as: 10t+/-4
Then, the perfect square must be equal to:
(10t+/-4)^2
= 100t^2 +/-80t +16
= 10t(10t+/-8) +16
This is equal to 
either:10t((10t+9) + 6
or: 10t(10t-7)+6
In each of these two cases, we observe that the tens digit is ODD.

We can now consider the only other form of the perfect square, that is, 10t+/-6
and arrive at the value of the perfect square as 
10t(10t+15)  +6  or, 10t(10t-9) 
so, the tens digit is ODD in this case as well.

Combining, the foregoing:
The tens digit of the perfect square must always be ODD.

By lemma -1, the perfect square must end in an even number of consecutive zeros.
Then, the nonzero digit preceding the first of these 0s must be  a 6.
So, the form of the perfect square must be:
abcd........06000......00
or abcd....6600...00
The tens digit then must be 0 or 6, which are even. 
This is a contradiction.
Consequently, there does NOT exist any perfect square having the said structure.

Edited on June 6, 2022, 2:14 am
  Posted by K Sengupta on 2022-06-06 01:50:38

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