Question

Sa se determine cel mai mic nr care impartit la 99da rest 51 , iar impartit la 77 da rest 29

Answer 1

n : 99 = x rest 51 => n = 99x + 51 | + 48

n : 77 = y rest 29 => n = 77y + 29 | + 48

=> n + 48 = 99 × ( x + 1)

=> n + 48 = 77 × (y + 1)


99 = 3^2 × 11

77 = 7 × 11
___________
cmmmc = 3^2 × 7 × 11 = 693


n + 48 = 693

n = 694 - 48

n = 645
________



645 : 99 = 6 rest 52

645 : 77 = 8 rest 29