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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!

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    $\begingroup$ No, because if $n$ has $k$ trailing zeros, $n^2$ has $2k$ trailing zeros. $\endgroup$ – Wojowu Feb 27 '19 at 17:17
  • $\begingroup$ @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. $\endgroup$ – Menotdan Feb 27 '19 at 17:21
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    $\begingroup$ 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. $\endgroup$ – Wojowu Feb 27 '19 at 17:27
  • $\begingroup$ 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 $\endgroup$ – Stan Tendijck Feb 27 '19 at 17:52
  • $\begingroup$ I am pretty new to square roots, so please try to explain some of the terms $\endgroup$ – Menotdan Feb 27 '19 at 17:59
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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.

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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 $\sqrt{105}\cdot\sqrt{10000}$

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    $\begingroup$ well, this doesn't really answer my question, but I guess this is cool? $\endgroup$ – Menotdan Feb 27 '19 at 18:20
  • $\begingroup$ @Menotdan when sqrt root of 10 is always decimal you can always separated from number with an odd number of tagging zero and you have even number of zero tagging wich is complete sqrt time to decimal and always been decimal i hope that explanation content you $\endgroup$ – Reza Feb 27 '19 at 18:26

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