Question

Gaseste numerele necunoscute:

967:(5a+1)+10rest7

(4b+3):47=19rest6

Answer 1

967:(5a+1)=10rest7⇒conform Teoremei impartirii cu rest: a = b ×q + r
 967=(5a+1)×10+7⇒
⇒967=10a+10+7
⇒967-17=10a
⇒950=10a
⇒a=950:10
a=95

(4b+3):47=19rest6 conform Teoremei impartirii cu rest: a = b ×q + r
⇒(4b+3)=47×19+6
⇒4b+3=893+6
⇒4b=899-3
⇒4b=896
⇒b=896:4
b=224