Question

Aflai numerele naturale abcd(cu bara) stiind ca abcd(cu bara)=ab*cd+abd (toate cu bara).Dau coroana

Answer 1

ab x cd+abd=abcd
ab x cd+10ab+d=100ab+cd
10ab+ab x cd -100ab=cd-d

1)  ab(cd-90)=10c

90≤cd≤99 pentru ca (cd-90)≥0
0≤cd-90≤9
din enunt punem conditia ca:
a≠0
c≠0
cd-90=0, c=0 nu e buna varianta 
cd-90=1, cd=91, c=9, d=1, ab=10c, ab=90, a=9, b=0  ⇒ abcd=9091
cd-90=2, cd=92, c=9, d=2, 2ab=10c, a=4, b=5  ⇒ abcd=4592
cd-90=3, cd=93, c=9, d=3, 3ab=10c, ab=30, a=3, b=0 ⇒ abcd=3093
cd-90=4, cd=94, c=9, d=4, 4ab=90 nu merge pentru ca ab e intreg
cd-90=5, cd=95, c=9, d=5, 5ab=90, ab=18, a=1, b=8, abcd=1895
cd-90=6, cd=96, c=9, d=6, 6ab=90, ab=15, a=1, b=5, abcd=1596
cd-90=7, cd=97, c=9, d=7, 7ab=90 nu merge
cd-90=8 nu merge, 8 nu divide pe 90
cd-90=9, cd=99, c=9, d=9, 9ab=90, ab=10, a=1, b=0, abcd=1099

la evaluarea fiecarei variante s-a tinut seama de relatia 1)
iar cd-d=10c+d-d=10c=c0 cu bara (e banal)

abcd={9091,4592,3093,1895,1596,1090}