Question

a) Determinati cel mai mic numar natural stiind ca impartit pe rand la 9 , 12 si 18 da restul 8 de fiecare data . 

b) Numerele 122 , 149 si 176 impartite la acelasi numar natural dau resturile respectiv 10 , 9 si 8 . Aflati impartitorul.

Answer 1

a)x:9=c1 rest 8
x=M₉ + 8

x:12=c2 rest 8
x=M₁₂ +8

x:18=c3 rest 8
x=M₁₈ +8

x=[M₉;M₁₂;M₁₈]+8
x-cel mai mic

9=3²
12=2² *3
18=2* 3²

[9;12;18]= 3² * 2²= 9*4=36

x=36+8= 44

2)122:x=c1 rest 10
122=xc1 +10
xc1=122-10
xc1=112

149:x=c2 rest 9
149=xc2 +9
xc2=149-9
xc2=140

176:x=c3 rest 8
176=xc3 +8
xc3=176-8
xc3= 168

x= (112;140;168)

112= 2⁴ *7
140= 2² * 5 *7
168=2³ *3 *7

x= 2² *7

x=4*7

x=28

Answer 2

a) Notam nr cautat cu a

a: 9 = b rest 8  ⇒ a=9b+8 ⇒ a-8=9b
a:12 = c rest 8 ⇒a=12c+8 ⇒a-8=12c
a:18 = d rest 8 ⇒a=18d +8 ⇒a-8=18d
 
=> 9b=12c=18d
cel mai mic multiplu comun al lui 9,12 si 18 =2*2*3*3=36
9=3*3
12=2*2*3
18=2*3*3

=> a-8=36
a=36+8= 44

b)Notam numarul cautat cu a

122: a=b  rest 10 ⇒ 122=ab +10 ⇒ab=112
149: a=c  rest 9⇒149=ac +9⇒ac=140
176: a=d  rest 8⇒ 176=ad +8⇒ad= 168

cautam un divizor comun  pt 112,140 si 168
112=2⁴*7
140=2²*5*7
168=2³*3*7
a ∈ {2,2²,7,2*7,2²*7}, adica
=> a ∈ {2,4,7,14,28}