Henry Ernest Dudeney/Puzzles and Curious Problems/131 - Feeding the Monkeys/Solution

by : $131$

 * Feeding the Monkeys
 * A man went to the zoo with a bag of nuts to feed the monkeys.
 * He found that if he divided them equally amongst the $11$ monkeys in the first cage he would have $1$ nut over;
 * if he divided them equally amongst the $13$ monkeys in the second cage there would be $8$ left;
 * if he divided them amongst the $17$ monkeys in the last cage $3$ nuts would remain.
 * He also found that if he divided them equally amongst the $41$ monkeys in all $3$ cages,
 * or amongst the monkeys in any $2$ cages,
 * there would always be some left over.


 * What is the smallest number of nuts that the man could have in his bag?

Solution

 * $2179$ nuts.

Proof
Let $n$ be the number of nuts he had in his bag.

We have the following congruences:

From $(1)$, we have that:
 * $n \in \set {1, 12, 23, 34, 45, 56, \ldots}$

From $(2)$, we have that:
 * $n \in \set {8, 21, 34, 47, 60, 73, \ldots}$

Thus:
 * $n \equiv 34 \pmod {11 \times 13}$

that is:
 * $n \equiv 34 \pmod {143}$

Thus:
 * $n \in \set {34, 177, 320, 463, 606, 749, 892, 1035, 1178, 1321, 1464, 1607, 1750, 1893, 2036, 2179, 2322, \ldots}$

From $(3)$, we have that:
 * $n \in \set {3, 20, 37, 54, 71, 88, 105, 122, 139, 156, 173, 190, 207, 224, 241, 258, 275, \ldots}$