4. Four men possess coins. If the first were to take 3 from the second, then he would have as many as would remain with the second. If, instead, the second were to take 1 from the first, then he would have three times as many as would remain with the first. How many coins do the two men possess in total?
(A) 14
(B) 18
(C) 12
(D) 13
(E) 16
(4) The correct answer is $(E)$.

Let $a$ and $b$ be the number of coins, respectively, of the first and the second. The problem states that the equations $b-3=a+3$ and $b+1=3(a-1)$ hold. From these, we get $b=a+6$ and $a+7=3(a-1)$, which simplifies to $2a=10$, or $a=5$ and $b=11$. In total, the two have $5+11=16$ coins. Problem proposed by A. Dal Zotto, S. Pelizzola, R. Zanotto.