Elements with order $100$ in $\mathbb{Q}/\mathbb{Z}$. How many elements with order $100$ are in $\mathbb{Q}/\mathbb{Z}$?
 A: Forty, I think: $\frac1{100}, \frac3{100}, \frac7{100}, \frac9{100}, \ldots, \frac{99}{100}$.
A: Hint. The order of $q\in\mathbb{Q}/\mathbb{Z}$ is the smallest natural number $n$ such that $nq\in \mathbb{Z}$. 
Count the numbers of the form $q=a/100$ with $a\in [1,100]\cap \mathbb{N}$ such that $\gcd(n,100)=1$. 
A: Building on the previous 2 answers, Euler's $\varphi$ function should equal the number of elements, so $\varphi(100)=2^2(1-1/2)(5^2)(1-1/5) = 4(1/2)(25)(4/5) = 40$
A: Consider the orders of $26/100$ and $27/100$ in this group:
$$
\begin{array}{clclc}
& 26\times 1 = 26 & & 27\times1=27 \\
& 26\times 2 = 52 & & 27\times2=54 \\
& 26\times 3 = 78 & & 27 \times 3 = 81 \\
& 26\times 4 = 104 \equiv 4 & & 27 \times 4 = 108 \equiv 8 \\
& 26 \times 5 = 130 \equiv 30 & & 27\times 5 = 135 \equiv 35 \\
& \qquad\vdots & & \qquad \vdots \\
\text{back to 0 after 50 steps }\longrightarrow & 26 \times 50 = 1300 \equiv0 & & 27 \times 50 = 1350 \equiv 50 \longleftarrow \text{not yet back to 0} \\
& & & 27 \times 51 = 1377 \equiv 77 \\
& & & 27 \times 52 = 1406 \equiv 6 \\
& & & 27 \times 53 = 1433 \equiv 33 \\
& & & \qquad \vdots \\
& & & 27 \times 100 = 2700 \equiv 0
\end{array}
$$
Every number will get you back to $0$ after $100$ steps; the reason $26$ gets you back there earlier is that it has the factor $2$ in common with $100:$
$$
\frac{26}{100} = \frac{2\times13}{2\times 50} = \frac{13}{50},
$$
so $26\times50 = (13\times2)\times50 = 13\times(2\times50) = 13\times100.$
Thus the ones that take the full $100$ steps to get back to $0$ are those that have no factors in common with $100$ except the trivial factor that all numbers share, which is $1.$
