Mangoes Riddle

Let the number of mangoes at the start be $n$ .

• The 1st customer bought $\dfrac n4 + 3 = p_1$ , where $p_1$ is a prime. The number of mangoes remain is $n_1 = n - \left(\dfrac n4+3\right) = \dfrac 34 n - 3$ .
• The 2nd customer bought $\dfrac {n_1}3 + 2 = \dfrac n4 + 1 = p_2$ , where $p_2$ is a prime. The number of mangoes remain is $n_2 = \dfrac 34n-3 - \left(\dfrac n4+1\right) = \dfrac n2 - 4$ .
• The 3rd customer bought $\dfrac {n_2}2 + 1 = \dfrac n4 - 1 = p_3$ , where $p_3$ is a prime. The number of mangoes remain is $n_3 = \dfrac n2-4 - \left(\dfrac n4-1\right) = \dfrac n4 - 3$ .

We note that $p_1 > p_2 > p_3$ and the difference between two consecutive primes is 2. Therefore, the three primes must be 3, 5 and 7.

That is $\begin{cases} p_1 = \dfrac n4 + 3 = 7 \\ p_2 = \dfrac n4 + 1 = 5 \\ p_1 = \dfrac n4 -1 = 3 \end{cases}$ $\implies n = 16$ and the number of mangoes left is $n_3 = \dfrac n4 - 3 = \boxed{1}$ .

Suppose the seller has $n$ mangoes left after all his sales.

• $n$ must be one less than half the number of mangoes he had before Customer #3, so he had $2n + 2$ mangoes before Customer #3.

• $2n + 2$ must be two less than two-thirds of the number of mangoes he had before Customer #2, so he had $3n + 6$ mangoes before Customer #2.

• $3n + 6$ must be three less than three-fourths of the number of mangoes he had before Customer #1, so he had $4n + 12$ mangoes before Customer #1.

Thus Customers #3, #2 and #1 bought $n + 2, n + 4$ and $n + 6$ mangoes respectively. It is known that the only three consecutive odd numbers which are prime are $3, 5$ and $7$ , so those must be the numbers of mangoes bought, which makes the number he has remaining $n = 1$ .

Each customer orders exactly 2 fewer mangoes than each previous one.

The only 3 prime numbers that each differ by 2 are 7,5,3.

So customer #1 bought 7 mangoes.

The original number of mangoes is (7-3)*4 = 16.

Total left = 16-7-5-3= 1 .

The answer is 1.