22) Three friends are dividing a basket of apples equally and notice they can do so without cutting the apples. A fourth friend arrives and they decide to redistribute the apples equally again, and they manage to do so without cutting the apples. One of the four goes home and eats 2 apples on the way. Upon arriving home, he realizes he can divide the remaining apples in half with his girlfriend without cutting them. What is the minimum number of apples that could have been in the basket originally?
(A) 12
(B) 16
(C) 24
(D) 36
(E) 50.
22) The answer is $\mathbf{( C )}$. Let $M$ be the number of apples. Since three friends can share the apples equally without cutting them, $M$ is a multiple of 3, i.e., $M=3k$ with $k$ a positive integer. Since the same operation can also be done with four, $M$ is also a multiple of 4, i.e., $M=12h$ with $h$ an integer. Each of the four friends will therefore take home $3h$ apples. The one who eats 2 on the way will end up with an even number of apples. We thus have $3h-2=2m$, from which $3h=2(m+1)$. Since 3 and 2 are coprime, the possible solutions are $h=2p, m+1=3p$, with $p$ an integer. The smallest positive solution is obtained for $p=1$: $m=2, h=2, M=24$.