6. Six men possess some coins. If the first takes 2 from the second, then he would have as many as the second would have left. If, instead, the second takes 1 from the first, then he would have three times as many as the first would have left. How many coins do the two men possess in total?
(A) 14
(B) 16
(C) 12
(D) 18
(E) 13
(6) The correct answer is $(C)$.

Let $a$ and $b$ be the number of coins, respectively, of the first and the second. The problem states that the equations $b-2=a+2$ and $b+1=3(a-1)$ hold. From these, we get $b=a+4$ and $a+5=3(a-1)$, which simplifies to $2a=8$, or $a=4$ and $b=8$. In total, the two have $4+8=12$ coins. Problem proposed by A. Dal Zotto, S. Pelizzola, R. Zanotto.