Răspuns :
Fie 2k+1 primul numar impar (cel mai mic).
Avem:
(2k+1)+(2k+3)+(2k+5)+(2k+7)=100-52
<=> 8k+(1+3+5+7)=48
<=> 8k+16=48
<=> 8k=32
<=> k=4
Deci numerele cautate sunt 2*4+1, 2*4+3, 2*4+5 si 2*4+7, adica 9, 11, 13 si 15.
Avem:
(2k+1)+(2k+3)+(2k+5)+(2k+7)=100-52
<=> 8k+(1+3+5+7)=48
<=> 8k+16=48
<=> 8k=32
<=> k=4
Deci numerele cautate sunt 2*4+1, 2*4+3, 2*4+5 si 2*4+7, adica 9, 11, 13 si 15.