Question

Aflati numerele prime a, b, c, d stiind ca 42a + 336b +42c + 126d = 2016

Answer 1

in expresia din enunt ambii membrii se pot impartii la 2
21a+168b+21c+63d=1008
168b+21(a+c+3d)=1008
168b este par, 1008 par deci obligatoriu suma (a+c+3d) e para si prin urmare:
1)  a,c,d nu pot fi simultan pare
2)  una din variabilele a,b,c sigur e para
3) b sigur e impar

b=3,5,7,11........
168b=168 x 3=504
168b=168 x 5=840
168b=168 x 7=1176 nu merge

b=3 ⇒ 21(a+c+3d)=1008-504=504 ⇒ a+c+3d=24
b=5 ⇒ 21(a+c+3d)=1008-840=168 ⇒ a+c+3d=8 nu merge

deci ramane doar varianta
b=3
a+c+3d=24
a=2, c+3d=22, c=7, d=5
c=2, a+3d=22, a=7, d=5
d=2, a+c=18, a=7, c=11
d=2, a+c=18, a=11, c=7

in final avem variantele:
a=2, b=3, c=7, d=5
a=7, b=3, c=2, d=5
a=7, b=3, c=11, d=2
a=11, b=3, c=7, d=2