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Ana are mere

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Posted

Ana si Maria au mere intregi.

Daca Maria ar avea de doua ori mai multe mere decat are acum si ar mai pune pe langa ele si merele Anei, ar avea in total 14 mere.

Daca fetele si-ar aduce prietene cate mere are Ana, fiecare dintre ele avand tot atatea mere cate are Ana si le-ar pune intr-un cos impreuna cu merele Mariei , ar avea in cos 21 de mere.

Cate mere are Ana, Cate mere are Maria?

Posted (edited)

Ana are 4 mere si Maria are 5 mere?

Sau solutie 2

Ana are 35/4 mere si Maria are -7/2 mere

E un sistem de 2 ecuatii cu 2 necunoscute

2*maria+x=14

maria+x*x=21

Cod sursa MATLAB

syms x
syms mama

[solutions_maria,solutions_x] = solve('2*maria+x=14','maria+x*x=21')

solutions_maria =

5
35/4


solutions_x =

4
-7/2

Offtopic: Thanks to MATLAB

Edited by gigaevil
Posted
Ana si Maria au mere intregi.

Daca Maria ar avea de doua ori mai multe mere decat are acum si ar mai pune pe langa ele si merele Anei, ar avea in total 14 mere.

Daca fetele si-ar aduce prietene cate mere are Ana, fiecare dintre ele avand tot atatea mere cate are Ana si le-ar pune intr-un cos impreuna cu merele Mariei , ar avea in cos 21 de mere.

Cate mere are Ana, Cate mere are Maria?

2M + A = 14

P x P + M = 21 <=> P^2 + M = 21

A = P

M - Maria

A - Ana

P - prietene

Maria are 5 mere

Ana are 4 mere

2 x 5 + 4 = 14

4^2 + 5 = 21

Posted

ana=a

maria=b

2b+a=14 => b=(14-a)/2

a*a+b=21 => a*a+(14-a)/2=21 => (dupa amplificare) 2*a*a+14-a=42 => 2*a*a-a=28

merele fiind nr intregi >=0 => din prima ecuatie ca a<=14 (in cazul in care maria ar avea minimul de 0 mere). deci a e cuprins intre 0 si 14

daca ar fi 0=> 2*0*0-0 != 28

daca ar fi 1=> 2*1*1-1 != 28

daca ar fi 2=> 2*2*2-2 != 28

daca ar fi 3=> 2*3*3-3 != 28

daca ar fi 4=> 2*4*4-4 = 28

deci a = 4, deci ana are 4 mere

2b+a=14=> 2b=10 => b=5 deci maria are 5 mere

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