Răspunsuri

2014-11-12T18:30:28+02:00
M = 4n + r    r∈{1,2,3,.......(n-1)}    2· (4n+r) + n = 148      8n +2r +n = 148  9n +2r = 148
9n = 148 - 2r  ⇒  9 divide (148 - 2r) ⇒ (148 -2r) ∈ (144, 135, 126 , 117,108,99.....9}
\ { 135,117,99, ...nr. impare... 27,9}  (148-2r) ∈{ 9·16. 9·14., 9·12.......... 9·2}
⇒ 2r ∈ {2,4,6,8,10,12,14,16}  ⇒  r ∈ {1,2,3,4,5,6,7,8} 
I. daca r=2    m=4n+2      2( 4n+2)+n =148    9n =144    n=16 ; m=66
II. daca  r=11    m= 4n+11        2(4n+11) +n = 148    9n= 126   n=14; m=66
           r=20   m=4n+20        2(4n+20)+n=148      9n= 108   n=12 m=66 dar r>n
⇒ perechile care indeplinesc conditiile problemei sunt: (m=66;n=16) si  (m=66,n=14)