Answer to Question #109177 in Algebra for Dr.K.V.Krishna Rao

Question

Four friends have collected certain no of mango fruits and decided to distribute them the next day. In the midnight one of them diveded fruits in to four equal groups.But one mango left and he has given to his animal.He consumed his share. After sometime second person did the same thing and one fruit has left and gave it to his servent.and taken his share of fruits.The third and fourth also did the same thing.and for them also one ,one each has left.Next day they sat together and distributed fruits equally and no fruit is left.what was the original no of fruits.

Answer 1

Assume total mango fruits as $L$ ,

$L_4$ is part the taken by the First friend,

$L_4=\frac{(L-1)}{4}\\$

Remaining part is $3L_4$ ,

$L_3$ ​ is part the taken by the Second friend,

$L_3=\frac{(3L_4-1)}{4}\\$

Remaining part is $3L_3$ ,

​ ​$L_2$ is part the taken by the Third friend,

$L_2=\frac{(3L_3-1)}{4}\\$

Remaining part is $3L_2$ ,

$L_1$ ​ is part the taken by the Fourth friend,

$L_1=\frac{(3L_2-1)}{4}\\$

Remaining part at the next day is $3L_1$ ,

$L_0=\frac{(3L_1)}{4}\\$

Here all the $L_0,L_1,L_2,L_3,L_4,L$ should be integers and it should be the only condition.

Trail and error method can be used to solve the problem.

So the small amount of mango fruits which satisfy the conditions are 765.

Other solutions are 1785, 2813, 3837, 4861, 5885, 6909, 7933, 8957, 9981, ...........

If there were 765 mangos initially, then 191 is one share of four persons. The 1st friend consumed a part, remaining part is 191×3=573 so, (573-1)/4=143 and it is an integer. That is the second friend consumed a part at midnight. Remaining part is 143×3=429, so (429-1)/4=107 and it is also an integer. Important part of this question is that a friend consumes only his share, another part (3 times of his share) stays as it is.