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The shortest prefix before the repeating tail of a decimal representation. For a reduced fraction with denominator $2^u{5}^{v}m$, where $m>1$ is coprime to ten, its length is $\max (u,v)$; the tail length is given by decimal period from multiplicative order.

The Fermat-Euler theorem states that, for a positive integer $q$ and an integer $a$ coprime to $q$, $a^{\phi (q)}\equiv 1(\mathrm{mod}q)$, where the Euler totient function $\phi (q)$ counts residue classes coprime to $q$. The case $q=1$ is trivial. For $q>1$, let $r_1,\dots ,r_{\phi (q)}$ be a reduced residue system. Multiplication by $a$ permutes it: equality of two resulting residues can be cancelled using the modular inverse of $a$, and the products remain coprime to $q$. Multiplying the residues before and after the permutation givesEvery factor on the right is a unit, so its product can be cancelled. This provesThe integer $359$ is prime: none of the primes $2,3,5,7,11,13,17$ up to $\sqrt{359}<19$ divides it. Therefore $\phi (359)=358$. Apply the theorem to $60$ and use the supplied congruence:To connect this to individual digits rather than merely assert periodicity, use long-division remainders $r_0=1$ andInduction gives $r_n\equiv 10^n(\mathrm{mod}359)$. Since $10$ is coprime to $359$, no remainder is zero. The congruence above therefore implies $r_{n+179}=r_n$ for every $n\ge 0$, using the unique representative in $1,\dots ,358$. In particularHenceThis is the mechanism of decimal period from multiplicative order: the period divides $179$.

The Fermat-Euler theorem states that for integers $m\ge 1$ and $a$ with $\gcd (a,m)=1$,where Euler's totient function $\phi (m)$ counts the residue classes coprime to $m$. The case $m=1$ is trivial. For $m\ge 2$, list a reduced residue system $r_1,\dots ,r_{\phi (m)}$. Multiplication by $a$ permutes these classes: each $a{r}_i$ is coprime to $m$, and $a{r}_i\equiv a{r}_j$ implies $r_i\equiv r_j$ since $a$ has a multiplicative inverse modulo $m$. Multiplying the permuted residues givesThe product is itself coprime to $m$ and can be cancelled using its inverse, proving the theorem.

The integer 359 is prime: none of the primes $2,3,5,7,11,13,17$, all the primes not exceeding $\sqrt{359}<19$, divides it. Therefore $\phi (359)=358$ and 60 is coprime to 359. Using the supplied congruence and the Fermat-Euler theorem,Now express the decimal expansion by long division. Put $r_0=1$ and for $n\ge 1$ defineThus $r_n\equiv 10^n(\mathrm{mod}359)$, and none of these remainders is zero. The congruence $10^{179}\equiv 1$ implies $r_{n+179}\equiv r_n$; since both are the unique representatives between zero and 358, they are equal. A digit is determined by the preceding remainder, namely $a_n=\lfloor 10r_{n-1}/359\rfloor$, soThis proves the required pure periodicity from the first digit. In fact its least period is 179: the multiplicative order of 10 modulo 359 divides the prime 179 and is not one, since $10\not\equiv 1(\mathrm{mod}359)$. The connection between the remainder recurrence and the period is the decimal period from multiplicative order.

A decimal representation with only finitely many nonzero fractional digits. A reduced fraction has this property exactly when its denominator has no prime factors other than two and five. The zero tail is distinguished from the genuine repeating block measured by decimal period from multiplicative order.