Five Minute Puzzle

Let $x$ be the total number of mangoes collected. The first friend takes $\frac{x}{3}$ and leaves $\frac{2x}{3}$ .

The second friend then divides the remaining mangoes into three piles, but with one mango left. Thus, each pile is $\dfrac{\frac{2x}{3} - 1}{3}$ and the second friend takes one of these piles for himself plus the extra mango, so he has $\dfrac{\frac{2x}{3} - 1}{3} + 1$ mangoes.

Because the number of mangoes each person received are equal, we can set $\frac{x}{3}$ equal to $\dfrac{\frac{2x}{3} - 1}{3} + 1$ and solve for $x$ .

$\frac{x}{3} = \dfrac{\left(\frac{2x}{3} - 1\right)}{3} + 1 \longrightarrow x = \left(\frac{2x}{3} - 1\right) + 3 \longrightarrow x = 6$

So the total number of mangoes collected is 6.

Let's start with the second friend. He divided the remaining mangoes into three piles with one left over, so let $x$ represent the number of mangoes in each of the piles he created. This means that he took $x + 1$ mangoes (since he took the leftover mango), leaving $2x$ mangoes behind for the last friend. Since everyone got the same number of mangoes, it must be that $x + 1 = 2x \longrightarrow x = 1$ .

This means that the second friend got $1 + 1 = 2$ mangoes and together the three friends collected a total of 6 mangoes.

To solve this problem I worked backwards.

The second friend splits the mangos that the first friend left into 3 parts, and there is one left over. He leaves 2 parts for the third friend and takes 1 part and the extra mango.

We know that they all end up with the same number of mangos so 2 parts will be the same as 1 part plus one mango. This tells us that the 'parts' must be 1 mango each. So each boy gets 2 mangos and there are six mangos total.

when the second friend came he took one third and one mango from the left and he had the same as the third

so (1m/3 + 1 = 1m/3 + 1m/3)

m = 3 , so the second and the third took 3+1=4 then the three friends had 3 mangoes

The second boy took one part and one remaining, hence, the division looks like this: x,x,x,1. the second boy must take x+1 and must be equal to the third boy, who took x+x, i.e., x+x=x+1, resulting to x=1. Since, the second and the third boy took 2 mangoes, by the assumption given in the problem, the first boy also took 2. Hence, the collected mangoes in the beginning were 6 mangoes.

Lets consider the total mangoes to be x. So the first person comes and get x/3 of the mangoes. Then the second person comes and sees that when dividing the mangoes by 3 there is a remainder of 1. So he takes (2x-3)/9 + 1 of these mangoes. Since they have got equal no. of mangoes , equating these two we get x=6. Therefore the total no. of mangoes is 6