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**Gusbob** 30) Let the total number of rabbits initially be n and the number deposited at each house is x.

**After the first house**, we have n-x rabbits. We cross the bridge to double the number we have. So we now have 2n - 2x rabbits

**After the second house**, we have (2n - 2x) - x rabbits. We cross the bridge to double the number we have. So we now have 2(2n - 3x) = 4n - 6x rabbits

**After the third house**, we have (4n - 6x) - x rabbits. We cross the bridge to double the number we have. So we now have 2(4n-7x) = 8n - 14x rabbits

**After the fourth house**, we have (8n - 14x) - x rabbits. We cross the bridge to double the number we have. So we now have 2(8n-15x) = 16n - 30x

**At the fifth house**, we have (16n-30x) - x rabbits, which equals 0 as we now have no more rabbits.

So 16n - 31x = 0. To find the minimum number of rabbits required, think about this logically:

the least amount of rabbits he can leave at each house is 1. So let x = 1. If x = 1, then 16n = 31. n = 31/16. But this is not a whole number, so x cannot equal 1.

Instead of checking every single integer up, we can look at why x = 1 doesn't work -> 31 is not divisible by 16. So if we look for the least common multiple of 16 and 31, we can solve for n easily.

__LCM of 31,16 = 496. Therefore we need 31x = 16n = 496 => (x = 16, n = 31)__

This can be extended to any number m houses. If you look at the pattern generated on each step, the equation for the amount of rabbits is:

$\displaystyle 2mn - (2^m - 2)x - x = 0 $

Where m = number of houses; n = initial number of rabbits; x = rabbits deposited at each house.