$A$ and $B$ can do a work in $13$ days and $26$ days, respectively. If they work for a day alternately, starting with $A$, then in how many days will the work be completed?

Time taken by $A=13$ days
Time taken by $B=26$ days

Let the total work be the Least Common Multiple (LCM) of $13$ and $26$:

$$\text{Total Work}=26\text{units}$$

Now, determine the individual efficiencies:

$$\text{Efficiency of }A=\frac{26}{13}=2\text{units/day}$$

$$\text{Efficiency of }B=\frac{26}{26}=1\text{unit/day}$$

They work alternately starting with $A$:

* Day 1 ($A$): $2\text{units}$
* Day 2 ($B$): $1\text{unit}$

Work completed in $1$ cycle of $2$ days:

$$2+1=3\text{units}$$

Number of complete $2$ -day cycles required to get close to $26$ units:

$$\lfloor \frac{26}{3}\rfloor =8\text{ cycles}$$

Work done in $8$ complete cycles ($16$ days):

$$8\times 3=24\text{units}$$

Remaining work:

$$26-24=2\text{units}$$

On the $17^{\text{th}}$ day, it is $A$ 's turn. Time taken by $A$ to finish the remaining $2$ units:

$$\text{Time}=\frac{\text{Remaining Work}}{\text{Efficiency of }A}=\frac{2}{2}=1\text{day}$$

Total days taken to complete the entire work:

$$\text{Total Time}=16+1=17\text{days}$$