Question

1).Cate nr. naturale de 3 cifre dau restul 7 la impartirea prin 20?
2).Suma a 2 nr. naturale nu depaseste 120.Aflati nr.,stiind ca prin impartirea celui mai mare la cel mai mic obtinem catul 9 si restul 10.

Va rog ajutati-ma!!!!!!!!!!!!!!!!!!!!!!!!!1

Answer 1

X:20=C REST 7
x=20c+7
x-7=20c+7-7
x-7=20c
x-7∈M20
M20={20,40,60,80,100,120,140,160,180....}
x-7=100⇒x=100+7⇒x=107
107:20=5 rest 7
x-7=120⇒x=127
127:20=6 rest 7
x-7=140⇒x=147
147:20=7 rest7
..................
x-7=980⇒x=987
987:20=49 rest 7
nr.sunt:107,127,147,167,187,207,227,247,267,287.......987
deci 50 de nr.de 3 cifre au prop mentionata in problema

2) a+b=120
a:b=9 rest 10
a=9b+10
9b+10+b=120
10b=120-10
b=110/10
b=11
a=120-b=120-11
a=109