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Ab barat - ba barat=27 sa se calculeze ab+ba+2 repede va roog

Răspuns :

[tex]\overline{ab}-\overline{ba} = 27 \\ 10a+b-(10b+a) =27 \\ 10a-10b+b-a = 27\\ 10(a-b)-(a-b) = 27 \\ (a-b)\cdot (10-1) = 27\\ (a-b)\cdot 9 = 27 \\ a-b = 27:9 \\ a-b = 3 \Rightarrow (a,b)= \Big\{(4,1);(5,2);(6;3);(7;4);(8,5);(9,6)\Big\} \\ \\ \\ \overline{ab}+\overline{ba}+2 = 10a+b+10b+a+2 =11a+11b+2 = \\\\=11(a+b)+2 =\\ \\ = \Big\{11(4+1)+2,\,11(5+2)+2,\,\dots,\,11(9+6)+2\Big\}=\\ \\ = \Big\{57,79,101,123,\,\dots,167\Big\} = \\ \\ =35+22k,\quad k = \overline{1,2,...,6}[/tex]

[tex]\overline{ab}-\overline{ba}=27\\ \\ 10a+b-(10b+a)=27\\ \\ 10a+b-10b-a=27\\ \\ 9a-9b=27\\ \\ 9(a-b)=27\\ \\ a-b=27:9 \\ \\ a-b=3 \Rightarrow a=b+3 \\ \\ val \ max \ a \ lui \ a \ este \ 9, \ deci \ val \ maxima \ a \ lui \ b \ este \ 9-3=6, \ iar \ val \ minima \ este \ 1\\ \\ \overline{ab}+\overline{ba}+2=10a+b+10b+a+2=11(a+b)+2=11(b+3+b)+2=11(2b+3)+2\\ \\ b\in \{1, \ 2, \ 3, \ 4, \ 5, \ 6 \} \Rightarrow 2b\in \{2, \ 4, \ 6, \ 8, \ 10, \ 12 \}\\ \\ \Rightarrow 2b+3 \in \{5, \ 7, \ 9, \ 11, \ 13, \ 15 \} \\ \\ \Rightarrow 11(2b+3) \in\{55, \ 77, \ 99, \ 121, \ 143, \ 165 \}\\ \\ \Rightarrow 11(2b+3)+2 \in \{57, \ 79, \ 101, \ 123, \ 145, \ 167 \}[/tex]