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.

#NumberTheory

Easy Math Editor

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Comments

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 $n^2 = 800...001$

Siddhartha Srivastava - 6 years, 3 months ago

Nice solution

A Former Brilliant Member - 6 years, 3 months ago

Nice solution :)

Calvin Lin Staff - 6 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$