Răspuns:
A∩B={2009}
Explicație pas cu pas:
A={xyzt cu proprietatea xyzt=100xy + 9},
B={abcd cu proprietatea abcd=69*ad + 10c +8}
xyzt=100*xy+9
xyzt=xy00+zt=xy*100+zt
xy*100+zt=100*xy+9
=> zt=9; z=0 și t=9
xy={10,11; 12; ......;99}
A={1009; 1109; 1209; 1309;...;2009; 2109; ....;9009; 9109;..;9909}
abcd=69*ad+10c+8
1000a+100b+10c+d=69*(10a+d)+10c+8
1000a-690a+100b=69d-d+10c-10c+8
310a+100b=68d+8
10(31a+10b)= 68d+8
=>u(68d+8)=0 => d={4;9}
d=4 => 10(31a+10b)=68*4+8
10(31a+10b)=280 /:10
31a+10b=28, nu convine
d=9 =>10(31a+10b)=68*9+8
10(31a+10b)=620 /:10
31a+10b=62 => a este nr par
=> a=2; b=0; c={0;1;2;….;9}; d=9
B={2009; 2019; 2029; 2039; ….;2099}
A∩B={2009}
