# What is the easiest way to find the $5$-th root?

A well known demonstration for number-theoretical properties is the following :

A two-digit-number is raised to the $5$ th power and the original number has to be found.

It can be shown with the Carmichael function or with easy modular arithmetic, that the ending digit of the $5$ power is the same as the ending digit of the original number. We have

$$n^5\equiv n\mod 10$$ for every integer $n$

To get the first digit, one method is to cancel the last $5$ digits and to find $k$ , such that $k^5$ is smaller , but $(k+1)^5$ is larger than the resulting number (In the case of the ending digit $0$, the last non-zero digit is the first digit of the original number).

I wonder whether the first digit can be determined faster and without memorizing the $5$-th powers of the numbers $1$ to $9$.

For example, does determining the given number modulo $9$ or modulo $11$ help ?

What is the easiest way to find the first digit ?

• The solution has to be determined without electronic help and without writing anything down on a paper. – Peter May 6 '17 at 19:13
• this number can be written in the form $$10a+b$$ and then? – Dr. Sonnhard Graubner May 6 '17 at 19:30
• $10 = 2\cdot 5$, so primitive root modulo 10 exists. 3 is one of them (four in total). If a given $n$ is coprime with 10, maybe you can associate $n$ with some power of 3 then mod 10 to get what you want quicker? – swoopin_swallow May 6 '17 at 19:55
• There are a few tools left !!!. – Felix Marin May 9 '17 at 2:48
• @FelixMarin True but if we allow, for example the notification of the $5$th powers upto $9^5$, it is trivial to determine the first digit. What I search is some easy rule (for example involving the digit sum or the number of digits) allowing to determine the first digit quickly and with a "in the head"-calculation. – Peter May 10 '17 at 10:07

Reduction modulo $11$ doesn't help much, unless the result turns out to be $0$ (then you immediately have the first digit: it's the same as the second), since $a^5 \equiv \pm 1 \pmod{11}$ if $11\nmid a$. A remainder of $+1$ or $-1$ would tell you whether $n$ is a quadratic residue modulo $11$, but that only reduces the options by half.
Reducing modulo $9$ is better. Since $\varphi(9) = 6$, we have $n^5 \equiv n^{-1} \pmod{9}$ if $3\nmid n$, so if $n$ wasn't divisible by $3$, that immediately tells you the remainder of $n$ modulo $9$, and thus - unless $0$ is a valid first digit - determines $n$. If $n$ was divisible by $3$, that leaves you with three options, but those three options should be fairly easy to distinguish.
But you can also reduce modulo $7$ - again, $n^5 \equiv n^{-1} \pmod{7}$ if $7 \nmid n$. That gives you the remainder of $n$ modulo $7$. And if the reduction modulo $9$ shows $3\mid n$, together with the remainder modulo $7$, the number is determined.