A logic problem by Victor Loh

x represents the amount of rabbits before the crossing of the first river. After crossing the river, the number of rabbits double, and he deposits "y" rabbits into the first house. Thus he has 2x - y rabbits left. Once he crosses the next river this quantity is doubled. Now he has 4x - 2y rabbits, and again deposits y" rabbits into the house, leaving him with 4x - 3y rabbits. This process is repeated until he reaches the fifth house, where he finally has 32x - 31y rabbits, which is given to be 0. Thus, since 31 is prime, x must be 31, and y must be 32. This is the smallest amount of rabbits that the magician can begin with to satisfy this problem, because the least common multiple between 31 and 32 involves both 31 and 32.

$x\\ 2x-y\\ 4x-3y\\ 8x-7y\\ 16x-15y\\ 32x-31y=0$

Thus, the minimum number of rabbits the magician can start with is:

$31$

Draw a diagram to visualize the problem as in the picture.

Picture

We will need to give $k$ rabbits at every house, so the number of rabbits the magician has is $k$ at the last house.

We can work backwards, and find that the number of rabbits the magician has when he starts is in the form $\frac{31k}{16}$ .

Since the number of rabbits is an integer and it must be the minimum, $k=16$ and $\frac{31\times16}{16}=\boxed{16}$

Suppose we have x rabbits in the beginning and we deposit y rabbits in every house. After deposit in the 1st house, we have x-y rabbits. After crossing 1st bridge we have 2(x-y) rabbits. After crossing 2nd bridge, we have 2(2(x-y)-y) rabbits and so on after deposit in the 5th house, we have the equation : 2(2(2(2(x-y)-y)-y)-y)-y=0. (because no rabbits left in the end) thus, 16x - 31y=0. Here, we must find the smallest value of x and y which are also integers. Here we find x= 31 and y=16. So, the number of rabbits in the beginning is 31.

let us suppose "x' no. of rabbits are given at each house and 'y' be the min. no. of rabbits he takes, then from last(5th) house after giving 'x' rabbits he is left with 0 rabbits , before crossing 5th river he must be having x/2 rabbits and he gave 'x' rabbits at 4th house so after crossing 4th river he would have 'x+x/2' ,i.e.,' 3x/2' rabbits , before crossing 4th river, he would have '3x/4' rabbits , after crossing 3rd river
he would have 3x/4+x = 7x/4 rabbits before crossing 3rd river, he would have '7x/8' rabbits , after crossing 2nd river he would have 7x/8+x = 15x/8 rabbits before crossing 2nd river, he would have '15x/16' rabbits , after crossing 1st river he would have 15x/16+x =31x/16 rabbits before crossing 1st river, he would have '31x/32' rabbits , 31x/32=y :x/y=32/31 for least no. of rabbits to be taken x &y must be co-primes so 'y' =31 rabbits