Question

Determinati toate numerele naturale de forma ab stiind ca:

a) 1ab+1ba=299
b) aabb+bbaa=3333
c) a0b+b0a=505
d) aba+bab=777

Answer 1

[tex]a)\overline{1ab}+\overline{1ba}=299\\ 100+10a+b+100+10b+a=299\\ 200+11a+11b=299\\ 11(a+b)=99\\ a+b=9;a,b \neq 0\\ S:(a,b)\in \{(1,8); (2,7);(3,6);(4,5);(5,4);(6,3);(7,2);(8,1)\}\\ \\ b)\overline{aabb}+\overline{bbaa}=3333\\ 1000a+100a+10b+b+1000b+100b+10a+a=3333\\ 1111a+1111b=3333\\ a+b=3;a,b\neq 0\\ S:(a,b)\in \{(1,2);(2,1)\}\\ \\ c)\overline{a0b}+\overline{b0a}=505\\ 100a+b+100b+a=505\\ 101a+101b=505\\ a+b=5; a,b\neq0\\ S:(a,b)\in \{(1,4); (3,2); (2,3); (4,1)\}\\ [/tex]
[tex]d)\overline{aba}+\overline{bab}=777\\ 100a+10b+a+100b+10a+b=777\\ 111a+111b=777\\ a+b=7; a,b\neq 0\\ S:(a,b)\in \{(1,6);(2,5);(3,4);(4,3);(5,2);(6,1)\}[/tex]

Answer 2

a) 145+154=299 a= 4 b=5 b) 1122+2211=3333 a=1 b=2 c) 203+302=505 a=2 b=3 d) 434+343=777 a=4 b=3