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determina n apartie lui N astfel incat (2n+3) divide (5n+12)
Dau CORONITA


Răspuns :

2n+3I5n+12
2n+3I2*(5n+12)
2n+3I10n+24
2n+3I2n+3
2n+3I5*(2n+3)
2n+3I10n+15
2n+3I(10n+24)-(10n+15)
2n+3I9
2n+3∈D9
D9={1,3,9}
Cazul 1:
2n+3=1⇒n=-1∉N
Cazul 2:
2n+3=3⇒n=0∈N
Cazul 3:
2n+3=9⇒n=3∈N
In concluzie:
{n}={0,3}
(2n+3)|(5n+12)⇒(2n+3)|[(5n+12)×2]⇒(2n+3)|(10n+24)

(2n+3)|(2n+3)
(2n+3)|[(2n+3)×5]⇒(2n+3)|(10n+15)

(2n+3)|(10n+24-10n+15)
(2n+3)|9⇒

⇒(2n+3)∈D₉
D₉={1, 3, 9}

(2n+3)∈{1, 3, 9}
(2n)∈{0, 6}
n∈{0, 3}