Question

a)determinati nr de forma ab pt care 2ab+ab3=830
b)determinati nr abc stiind ca ab3+a5c+7bc+1243
c)determinati nr de forma ab astfel incat 21ab+ab43
3860
d)determinati nr de forma ab stiind ca 4 ori ab+4 ori ba=176
VA ROG AJUTOR111
DAU CORONITA!

Answer 1

a). 2ab+ab3=830
200+10a+b+100a+10b+3=830
110a+11b=830-203
11(10a+b)=627
10a+b reprezinta numar de forma ab
ab=627:11
ab=57 => a=5 si b=7

b). ab3+a5c+7bc+1243 (banuiesc ca e ceva gresit si forma corecta este ab3+a5c+7bc=1243)

100a+10b+3+100a+50+c+700+10b+c=1243
200a+20b+2c=1243-753
2(100a+10b+c)=490
numarul de forma abc=490:2=245 => a=2, b=4, c=5

c). 21ab+ab43=3860

2000+100+10a+b+1000a+100b+40+3=3860
1010a+101b=3860-2143
101(10a+b)=1717
numarul de forma ab=1717:101=17 => a=1 si b=7

d). 4(ab+4)*ba=176 
(10a+b+4)*(10b+a)=176:4=44