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Cate numere de forma 2xy5 ,in baza zece , sunt divizibile cu 7.

Răspuns :

7/2xy5 <=> 7/(2000+100*x+10*y+5)<=>7/(7*286 +3+7*14*x+2*x+7*y+3*y)<=>
<=>7/(2*x+3*y+3)
1)daca x=0=> 7/(3*y+3) <=>y=6
2)daca x=1 =>7/(3*y+5) <=>y=3
3)x=2=>7/(19+3*y) <=>7/(5+3*y)<=>y=3
...........si verifici pana la x=9