Question

Primele 3 exerciții incercuite

Answer 1

x:16=c1r6
x:20=c2r10
x:24=c3r14
111<x<502
scriu teorema impartirii cu rest
x=16c1+6=16c1+16-10=16(c1+1)-10
x=20c2+10=20c2+20-10=20(c2+1)-10
x=24c3+14=24c3+24-10=24(c3+1)-10
adun 10
x+10=16(c1+1)=20(c2+1)=24(c3+1)
x+10 e multiplu comun al nr 16, 20 si 24 si x trebuie sa fie in intervalul (111,502)
calculez [16,20,24]
16=2^4
20=2^2*5
24=2^3*3
[16,20,24]=2^4*3*5=240
x+10 ∈ {240,480}
x∈{230,470}

7. La ex acesta faci la fel observand ca diferenta intre impartitor si rest e 3
vei obtine ca x+3 e [10,12,15] 
10=2*5
12=2^2*3
15=3*5
[10,12,15]=2^2*3*5=60
x=57

6.   x:6=c1r5
      x:7=c2 r5
x=6c1+5=7c2+5
x-5=6c1=7c2
x-5 e multiplu al nr 6 si 7 care respecta conditia x:17=c3r12, x=17c3+12
x-5∈{42,84,126,168,210......}
x∈{47,89,131,173,215.....}
Scad 12 si verific care nr e multiplu de 17
M17={17,34,51,68,85,102,119}
Observ ca 131-12=119 care e multiplu de 17
Rezulta ca nr cautat este 131
v:   131:6=21 r5
       131:7=18 r5
       131:17=7 r12