First I'll show that series resistances add. Let's take one branch of 10 Ω + 20 Ω + 30 Ω resistors. The current through each of them is the same, let's give it a name: \$I\$. Then according to Ohm's Law we have the following voltages across the resp. resistors:
\$ (I \times 10 \Omega) \$ across the \$ 10 \Omega \$ resistor,
\$ (I \times 20 \Omega) \$ across the \$ 20 \Omega \$ resistor,
\$ (I \times 30 \Omega) \$ across the \$ 30 \Omega \$ resistor,
for a total voltage of
\$ V = (I \times 10 \Omega) + (I \times 20 \Omega) + (I \times 30 \Omega) = I \times (10 \Omega + 20 \Omega + 30 \Omega) \$
so that
\$ R_{total} = \dfrac{V}{I} = (10 \Omega + 20 \Omega + 30 \Omega) = 60 \Omega \$
The total resistance of resistors in series is the sum of their resistances.
So we can simplify to three 60 Ω resistors in parallel. This time we won't have the same currents but the same voltage across each resistor. Then the currents through the branches are:
\$ \dfrac{V}{60 \Omega} \$ through the first \$ 60 \Omega \$ resistor,
\$ \dfrac{V}{60 \Omega} \$ through the second \$ 60 \Omega \$ resistor,
\$ \dfrac{V}{60 \Omega} \$ through the third \$ 60 \Omega \$ resistor,
for a total current of
\$ I = \dfrac{V}{60 \Omega} + \dfrac{V}{60 \Omega} + \dfrac{V}{60 \Omega} = V \times \left(\dfrac{1}{60 \Omega} + \dfrac{1}{60 \Omega} + \dfrac{1}{60 \Omega} \right)\$
so that
\$ \dfrac{I}{V} = \dfrac{1}{R} = \dfrac{1}{60 \Omega} + \dfrac{1}{60 \Omega} + \dfrac{1}{60 \Omega} \$
and thus
\$ R = \dfrac{1}{\dfrac{1}{60 \Omega} + \dfrac{1}{60 \Omega} + \dfrac{1}{60 \Omega}} = \dfrac{1}{\left(\dfrac{3}{60 \Omega}\right)} = \dfrac{60 \Omega}{3} = 20 \Omega\$
The equivalent resistance of \$N\$ resistors of \$R\$ Ω each is \$R/N\$ Ω.
In practice you'll often have two different resistors in parallel, then you have to use the equation I mentioned earlier:
\$ R = \dfrac{1}{\dfrac{1}{R1} + \dfrac{1}{R2}} = \dfrac{1}{\left(\dfrac{R1 + R2}{R1 \times R2}\right)} = \dfrac{R1 \times R2}{R1 + R2} \$