Question

1.se dau numerele naturale a si b astfel incat (a, b)=10 si [a, b]=60
determinati numerele
2.determinati cel mai mic numar natural de trei cifre, care impartit la 14,21 si 28 da restul 13.
VA ROG CAT MAI RAPID PUTETI DAU COROANA
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Answer 1

Răspuns:

Explicație pas cu pas:

a*b=60*10=600

a=10x si b=10y  s x si y prime intre ele

10x*10y=60

xy=6 si x si y prime intre ele

convin 1, 6 si 2,3

deci

a=10...b=60

sau

a=20  si b=30

2

n= [14,21,28]k=13=7[2,3,4]k+13=7[3.4]k+13=7*12k+13=84k+13

84k+13≥100 ...84k≥87

84(k-1)+13<100

convine k=2

84*1+13=97<100

84*2+13=181

Answer 2

1.

$\it a\cdot b=(a,\ b)\cdot\[[a,\ b]=10\cdot60=600\ \ \ \ \ (1)\\ \\ \\ (a,\ b)=10 \Rightarrow \begin{cases} \it a=10x\\ \\ \it b=10y\\ \\ \it (x,\ y)=1\end{cases}\ \ \ \ \ (2)$

$\it (1), (2) \Rightarrow 10x\cdot10y=600|_{:100} \Rightarrow x\cdot y=1\cdot6=2\cdot3=3\cdot2=6\cdot1\ \ \ \ (3)\\ \\ (2),\ (3) \Rightarrow \begin{cases}\it a=10\cdot1=10;\ \ b=10\cdot6=60\\ \\ \it a=10\cdot2=20;\ \ b=10\cdot3=30\\ \\ \it a=10\cdot3=30;\ \ b=10\cdot2=20\\ \\\it a=10\cdot6=60;\ \ b=10\cdot1=10 \end{cases}$

2.

Notăm numărul cerut cu n.

$\it n-13\in M_{14}\cap M_{21}\cap M_{28}\\ \\ 4=2\cdot7\\ \\ 21=3\cdot7\\ \\ 28=4\cdot7\\\rule{110}{0.6}\\\[[14,\ 21,\ 28]=2\cdot3\cdot4\cdot7=168\\ \\ n-13=168|_{+13}\Rightarrow n=181$