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

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!

• No, because if $n$ has $k$ trailing zeros, $n^2$ has $2k$ trailing zeros. Commented Feb 27, 2019 at 17:17
• @Wojowu hmm it appears that is true, but how does that correspond to my question? I can kind of see it, but the connection is not quite clicking in my head. Commented Feb 27, 2019 at 17:21
• If a number has whole square root, it's of the form $n^2$. This means that its number of trailing zeros is $2k$, so even. Commented Feb 27, 2019 at 17:27
• This does not fully answer the question. You also need to proof that there can't be any extra zeros appearing in the square. Hint for proving that: try to find prime factors of the $x$ for which that happens by assuming that we can find an $x$ which satisfies that property Commented Feb 27, 2019 at 17:52
• I am pretty new to square roots, so please try to explain some of the terms Commented Feb 27, 2019 at 17:59

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