Question

La un centru de paine vin pe rand trei cumparatori: primul cumpara jumate din painea existenta plus o jumate de paine, al doilea cumpara o jumate din painea ramasa si inca o jumatate de paine, iar al treilea, la fel, cumpara o jumate din painea ramasa si inca o jumate de paine. Stiind ca dupa al treilea cumparator nu a mai ramas paine in centru, atunci cate paini au fost la inceput?

Answer 1

x=nr paini
x/2+1/2= (x+1)/2=1 cumparator
x-x/2-1/2=(x-1)/2 =rest
[(x-1)/2 ]/2+1/2= (x-1)/4+1/2 =x/4-1/4+1/2=x/4+1/4=(x+1)/4=al 2-lea cumparator
(x-1)/2-(x+1)/4=(2x-2-x-1)/4=(x-3)/4 =rest
[(x-3)/4 ]/2+1/2=(x-3)/8+1/2=x/8-3/8+1/2=x/8-1/8=(x+1)/8=al 3-lea cumparator
(x+1)/2+(x+1)/4+(x-1)/4=x
2x+2+x+1+x-1=4x
7x+7=8x
x=7 paini

1 ia 7/2+1/2=3,5+0,5=4 paini
7-4=3 paini rest
al 2-lea ia 3/2+1/2=1,5+0,5=2 paini
3-2=1 paine rest
al 3-lea ia  1/2+1/2=0,5+0,5= 1 paine