a) 5000+100a+10b+c=2×(1000a+100b+10c+5) +1190
5000+100a+10b+c= 2000a+200b+20c+10+1190
5000- 1190-10=2000a-100a+200b-10b+20c-c
3800= 1900a+190b-19c
3800=19×(100a+10b+c) | ÷19
200=100a+10b+c
a=2, b=0,c=0 deci numarul este 200
b) 1000a+100b+20+4=2×(1000+200+10a+b)+1054
1000a+100b+24=2000+400+20a+2b+1054
1000a-20a+100b-2b=2400+1054-24
980a+98b=3430
98×(10a+b)=3430 | ÷98
10a+b=35 deci numarul este 35
c) 7×(10a+b)+10b+a=(1000a+100b+10b+a)÷11+1
70a+7b+10b+a=(1001a+110b)÷11+1
71a+17b=11×(91a+10b)÷11+1
71a+17b= 91a+10b+1
17b-10b=91a-71a+1
7b=20a+1
b=3 , a=1
7×3=20×1+1
7×3=21 deci numarul este 13
d) ab + bc + ca = abc
(ab + ca) + bc = 100 a + bc
ab + ca = 100 a
(10a + b ) + (10c + a) = 100 a
b + 10 c = 89 a
a = 1 de unde b + 10 c = 89 de unde b = 9 si c = 8
verificare : 19+91+18=198
e) la unitati vom avea a+a+a=3 de unde a=1
la zeci vom avea 1+b+c=1b de unde 1+c=10 de unde c=9
si 1+b+9=1b deci b poate lua orice valoare de la 1 la 9
