16. Presupuneam ca cel mai mare nr dintre cele 10 este mai mic sau egal cu 10.
In acest caz cele 10 nr ar fi:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20.
Suma lor este 2+4+6...+20= 2(1 + 2 + ..10) = 2( 1+10) × 10/2 = 11/10 = 110
Dar in ipoteza stim ca suma trebuie sa fie 112 , 112>110.
Deducem ca cel mai mare nr este mai mare decat 20.
15. 2 + 4+ 6 + 8 + 10 + 12 + 14+ 16 + 18 + 20= 5×20+10= 110
110>108 rezulta ca 2 nr sunt egale.
2 + 4+ 6 + 8 + 10 + 12 + 14 + 16 + 18 +18 = 108 sau
2+2+6+8+10+12+14+16+18+20 = 108 sau oricare nr.
17. ( n+1) + ( n + 3) + ( n +5) + ( n+ 7) + ( n + 9) + (n + 11) + ( n + 13) + ( n + 15) + ( n+ 17) + ( n+ 19) = 100
10n =100 - ( 1+3+5+7+9+11+13+15+17+19)
10n = 100 - 100
n = 0
Atunci nr vor fi:
(n+1)+(n+3)+(n+5)+(n+7)+(n+9)+(n+11)+(n+13)+(n+15)+(n+17)+(n+19)
(n+1)= 0+1=1
(n+3)= 0+3=3
(n+5)=0+5=5
(n+7)=0+7=7
(n+9)=0+9=9
(n+11)=0+11=11
(n+13)=0+13=13
(n+15)=0+15=15
(n+17)=0+17=17
(n+19)=0+19=19
(1,3,5,7,9,11,13,15,17,19)
18. (2a+1)+(2a+3)+(2a+5)+(2a+7)+(2a+9)+(2a+11)+(2a+13)+(2a+15)+(2a+17)+(2a+19)=96
20a+(1+3+5+7+9+11+13+15+17+19)=96
2a+100=96
20a=4
dintr-un nr trebuie scazut 4 adica nr pot fi 3, 5 (7-4=3)
1,3,5,7,9,11,13,15,17,19,21
si putem obtine si alte 10 nr. prin scaderea lui patru din celelalte
9-4=5 3,5 (7-4=3) 1,3,5,7,9,11,13,15,17,19,21
11-4=7 3,5 , 7,9,11,13,15,17,19,21
21-4=17 3,5,7,9,11,13,15,17
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19. (2a+1)+(2a+3)+(2a+5)+(2a+7)+(2a+9)+(2a+11+(2a+13)+(2a+15)+(2a+17)+(2a+19)=120
20a+(1+3+5+7+9+11+13+15+17+19)=120
20a+100=120
20a=20
a=1
Adica nr pot fi: 3,5,9,11,13,15,17,19,21 suma=120
21>20
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20. o suma de nr pare , dar tot nr par , deci raspunsul este negativ
