Question

Aflati cel mai mic nr . natural care inmpartit pe rand la numerele 24 , 40 si 58 , da acelasi rest 5 si caturile nenule .

Answer 1

D:I=C,+R
R<I
R<2
R-nr natural        =>R=1

x:5=c1 rest 1
x=M₅ +1
scuzama dar cum adica 
x:7=c2 rest 1
x=M₇+1

x:2=c3 rest 1
x=M₂+1

x=[M₅;M₇;M₂]+1
x-cel mai mic

x=5*7*2+1
x=35*2+1
x=70+1
x=71

Answer 2

D=IxC+R;  R<I
n:24=C1+5
n:40=C2+5
n:58=C3+5
n=24xC1+5
n=40xC2+5
n=58xC3+5
n-5=24C1
n-5=40C2
n-5=58C3
[24,40,58]=c.m.m.m.c=2 la a 4x3x29=1392
n-5=1392+5
n=1397
deci nr este 1397