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.

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TopNewestNot 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 \( n^2 = 800...001 \) – Siddhartha Srivastava · 1 year, 7 months ago

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– Calvin Lin Staff · 1 year, 7 months ago

Nice solution :)Log in to reply

– Azhaghu Roopesh M · 1 year, 7 months ago

Nice solutionLog in to reply

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– Azhaghu Roopesh M · 1 year, 7 months ago

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 :)Log in to reply

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– Azhaghu Roopesh M · 1 year, 7 months ago

I'm sorry but I'm not getting what you are trying to sayLog in to reply

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– Azhaghu Roopesh M · 1 year, 7 months ago

Ok .Log in to reply