10).
A=17a+17b+25 : 17
17a:17=a rest 0
17b:17=b rest 0
25:17=1 rest 8 ( verificare cu formula: d=î*c+r , 25=17*1+8 )
0+0+8= 8 ( restul impartirii )
B=16a+28b+13 : 4
16a:4=4a rest 0
28b:4=7b rest 0
13:4 = 3 rest 1 ( verificare: d=î*c+r , 13=4*3+1 )
0+0+1= 1 (restul impartirii)
11).
x:30=c, rest 8
y:35=c, rest 34
(3x+2y):10 = c,rest ?
----------------------------
d=î*c+r, in cazul de fața: x=30*c+8
y=35*c+34
3(30*c+8) + 2(35*c+34)=90c+24+70c+68=160c+92
(160c+92):10=c,rest ?
160c:10=16c rest 0
92:10=9 rest 2
0+2=2 ( restul lui 3x+2y impartit la 10 )
12).
a+b+c=232(ecuatia 1)
a:b=14 rest 5(ecuatia 2)
b:c=7 rest 1(ecuatia 3)
a,b,c=?
-------------------
d=î*c+r
a=b*14+5 ( inlocuiesc a in ecuatia 1)
b=c*7+1 ⇒ c=(b-1)/7 (inlocuiesc c in ecuatia 1)
(b*14+5)+b+(b-1)/7=232
14b+5+b+(b-1)/7 = 232 ( numitor comun pe 7)
98b+35+7b+b-1=1624
106b+34=1624
106b=1590
b=1590:106
b=15
a=b*14+5=210+5=215
c=(b-1)/7=(15-1)/7=14:7=2
215+15+2=232
13).
a+b+c=297
a:b=2 rest 25
a:c=11 rest 8
a,b,c=?
---------------
d=î*c+r
a=b*2+25 ⇒b=(a-25)/2
a=c*11+8 ⇒c=(a-8)/11
(am scos b si c in functie de a,apoi inlocuiesc in prima ecuatie si calculez)
a+(a-25/2)+(a-8)/11=297 ( factor comun pe 22)
22a+11a-275+2a-16=6534
35a-291=6534
35a=6825
a=6825:35
a=195
b=(195-25)/2=170/2=85
c=(195-8)/11=187/11=17
195+85+17=297
