Counting Stones

Question

If you have a bucket of stones and remove two at a time, one will be left. If you remove three at a time, two will be left. If they're removed four, five, or six at at time, then three, four, and five stones will remain. If they're removed seven at a time, no stones will be left over. What is the smallest possible number of stones that could be in the bucket? How do you know?

Answer 1

Based on the comment below, I am adding a bit more honesty to this response:

The conditions of the problem are to find $n$ such that $n \equiv 1 \text{ (mod 2)}$, $n \equiv 2 \text{ (mod 3)}$, $n \equiv 3 \text{ (mod 4)}$, $n \equiv 4 \text{ (mod 5)}$, $n \equiv 5 \text{ (mod 6)}$, $n \equiv 0 \text{ (mod 7)}$.

$n \equiv 1 \text{ (mod 2)} \wedge n \equiv 2 \text{ (mod 3)} \Rightarrow n \equiv 5 \text{ (mod 6)}$, so the first two conditions don't really help us get anywhere. Next, $n \equiv 5 \text{ (mod 6)} \wedge n \equiv 3 \text{ (mod 4)} \Rightarrow n \equiv 11 \text{ (mod 12)}$. Lastly, $n \equiv 11 \text{ (mod 12)} \wedge n \equiv 4 \text{ (mod 5)} \Rightarrow n \equiv 59 \text{ (mod 60)}$, since the number has to end in a 4 or a 9 and also 11, 23, 35, 47, 59 are candidates. So we need to find that the smallest $n$ which is divisible by 7 such that $n \equiv 59 \text{ (mod 60)}$. The second possible choice is 119.

First, I had solved it computationally:

two_store = vector() three_store = vector() four_store = vector() five_store = vector() six_store = vector() for (i in 1:100){ cur = 7*i; if (cur%%2 == 1){ two_store = c(two_store, cur); } if (cur%%3 == 2){ three_store = c(three_store, cur); } if (cur%%4 == 3){ four_store = c(four_store, cur); } if (cur%%5 == 4){ five_store = c(five_store, cur); } if (cur%%6 == 5){ six_store = c(six_store, cur); } } two_three = intersect(two_store,three_store) four_five = intersect(four_store,five_store) two_five = intersect(two_three, four_five) total = intersect(two_five,six_store)

Answer 2

If there had been just one more stone in the bucket, the number of stones would have been a multiple of 2, 3, 4, 5, and 6, and therefore a multiple of the least common multiple of 2, 3, 4, 5, and 6, which is 60. Thus the number of stones is one less than a multiple of 60, or in the sequence 59, 119, 179, etc. The smallest number of 7 in that sequence is 119.