Question

bunica a dat mere la 4 nepoti
1=a primit o treime+32 mere
2=a primit o treime din rest+32
3=a primit o treime din ce i-a ramas+32
4= a primit o treime din ce a ramas si inca ultimele 32

cate mere a primit fiecare?

Answer 1

bunica da celor 4 nepoti ai ei, in total, "a" mere:
nepot1= (1/3)*a + 32;
rest1=a - nepot1= a - [(1/3)*a + 32]= a-(1/3)*a-32= (2/3)*a-32

nepot2=(1/3)*rest1 + 32 =(1/3)*[(2/3)*a-32] + 32 = (2/9)*a - 32/3 + 32 = (2/9)*a + 64/3;  
 rest2=rest1- nepot2= (2/3)*a-32 - [(2/9)*a + 64/3] = (2/3)*a-32 - (2/9)*a -64/3= (6/9)*a-(2/9)*a-96/3-64/3= (4/9)*a - 160/3

nepot3=(1/3)*rest2 + 32 =(1/3)*[(4/9)*a - 160/3] +32= (4/27)*a - 160/9 + (32*9)/9= (4/27)*a - 160/9 + 288/9= (4/27)*a +128/9;
rest3=rest2- nepot3=(4/9)*a - 160/3 - [(4/27)*a +128/9] = (4/9)*a - 160/3 - (4/27)*a -128/9 = (12/27)*a - (4/27)*a - 480/9 - 128/9 = (8/27)*a - 608/9

nepot4=(1/3)*rest3 + 32  (*)  
nepot4=a-(nepot1 + nepot2 + nepot3) (**)
din relatiile (*) si (**) prin tranzitivitate, rezulta: 

(1/3)*rest3 + 32 = a-(nepot1 + nepot2 + nepot3)
sau prelucrand:
(1/3)*[(8/27)*a - 608/9]+32=a-[(1/3)*a + 32 + (2/9)*a + 64/3 + (4/27)*a +128/9] --> (8/81)*a - 608/27 + (32*27)/27 = a - [(19/27)*a + 608/9] -->
(8/81)*a - 608/27 + 864/27= a- (19/27)*a - 608/9 --> (8/81)*a + 256/27 = (8/27)*a - 608/9 --> 608/9 + 256/27= (8/27)*a - (8/81)*a
--> (24/81)*a - (8/81)*a = 1824/27 + 256/27 --> (16/81)*a= 2080/27 --> a = (2080/27) : (16/81) --> a = (2080/27) * (81/16) = 130 * 3 =390

ieeee! Bunica are 390 de mere! 
PS am avut ceva emotii ca n-o sa-mi dea un numar natural! :)))

:D

Deci fiecare nepot primeste asa:
nepot1= 162 mere
nepot2= 108 mere
nepot3= 72 mere
nepot4= 48 mere