Is there a number that has an odd number of tagging zeros, but has a whole square root?

Question

I was doing some math, and I came across the problem $\sqrt{1000}$ And I was thinking about other square roots of numbers with various amounts of zeros. And it occurred to me that it seemed like numbers with odd numbers of tagging zeros had a decimal square root.

So I was curious: Is there a number with an odd number of tagging zeros, but has a whole number for its square root?

For example: 102000 has three tagging zeros, but its square root is 319.374388453

Note: I am new to square roots, so please try to explain some of the terms to me.

Thanks!

Answer 1

The number $n > 0$ has the form $n = 2^k \cdot 5^l \cdot r$, where $k, l$ are non-negative integers and $r$ is a positive integer not divisible by $2$ and $5$. Let $m = \min(k,l)$. Then $n$ has exactly $m$ tagging zeros because each tagging zero corresponds to a factor $10 = 2\cdot 5$. Let $n$ have an integer square root $s$. Write $s = 2^{k'} \cdot 5^{l'} \cdot r'$ as above. Then $n = s^2 = 2^{2k'} \cdot 5^{2l'} \cdot (r')^2$. Since $r'$ is not divisible by $2$ and $5$, also $(r')^2$ is not divisible by $2$ and $5$. We conclude $k = 2k'$ and $l = 2l'$, hence $m = \min(2k',2l') = 2 \min(k',l')$ is even.

Answer 2

If you have a number with an odd number of tagging zeros you can always separate the square root and you will have a decimal number times another decimal number

For example:

For $\sqrt{10500000}$ You can write $\sqrt{105}\cdot\sqrt{10000}$