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# Strings of Zeros

81 is a square number. Find the minimum number of zeros needed to be inserted between 8 and 1 such that it makes another perfect square, or prove that it is impossible.

Note by Bryan Lee Shi Yang
2 years, 9 months ago

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Not possible.

Suppose there does exist some $$n$$ where $$n^2 = 800...001$$. Then since $$800..001$$ is odd, $$n$$ must. Therefore, $$n = 2k + 1$$. Then,

$$(2k + 1)^2 = 800...001$$

$$4k^2 + 4k + 1 = 800..001$$

$$4k(k+1) = 800...001$$

$$k(k+1) = 200...000$$

$$k(k+1) = 2 * 2^t*5^t$$ for some $$t > 1$$

$$k(k+1) = 2^{t+1}*5^t$$

Now, $$\gcd(k,k+1) = 1$$.Also, since $$5^t > 2^{t+1}$$, we must have $$k+1 = 5^t$$ and $$k = 2^{t+1}$$. But, we can easily prove that $$5^t > 2^{t+1} + 1$$ for $$t > 1$$. Therefore, we have a contradiction, and there exist no integer $$n$$ for which $$n^2 = 800...001$$

- 2 years, 9 months ago

Nice solution :)

Staff - 2 years, 8 months ago

Nice solution

- 2 years, 9 months ago

Comment deleted Mar 13, 2015

Sorry , but did you mean that you salute me ? It's actually Siddhartha who posted the solution , so I guess you should be saluting him :)

- 2 years, 9 months ago

Comment deleted Mar 13, 2015

Comment deleted Mar 13, 2015

I'm sorry but I'm not getting what you are trying to say

- 2 years, 9 months ago

Comment deleted Mar 13, 2015

Ok .

- 2 years, 9 months ago