Question

x^10 +y(3^z +1)=2000, x,y,z-cifre

Answer 1

x=0
y(3^z+1)=2000
Y=4 ...3^z+1=500...3^z=499 imposibil
y=5  ...3^z+1=400...   3^z=399 imposibil

x=1
y(3^z+1)=1999
1999 nu e divizibil cu 2,3,4,5,6,8 sau 9
verificam cu 7..1999:7=nu
y=1
3^z+1=1999
3^z=1998 imposibil in N

x=2
x^10=1024

y(3^z+1)=2000-1024=976

y=1
3^z+1=976
3^z=975 exclus

y=2
3^z+1=976:2=488
3^z=478 imposibil in N

y=4
3^z+1=976:4=244
3^z=243 z=5, solutie

y=8
3^z+1=122
3^z=121 imposibil


y=9
3^z+1=976:9=nu se imparte

deci solutie unica
x=2
y=4
z=5

verificare
2^10+4(3^5+1)=1024+4(243+1)=1024+4*244=
=1024+976=2000
adevarat , bine rezolvat