Question

Aflati multimea divizorilor numerelor:
a=1+2+3+....+21
b=2+4+6+....+12
c=15+18+....63
d=17+21+....+41

Answer 1

a=1+2+3+....+21

a = 21(1+21): 2= 21.11= 231;

a : 1, 3, 7, 11, 231;

b=2+4+6+....+12

b= 2( 1 + 2 +3 + 4 + 5 + 6) = 2.6.(1 + 6):2 = 42;

b: 1, 2, 3, 6, 7, 21, 42;

c = 15 + 18 +.... 63;

c = 3. (5 + 6 + 7 + ...  + 21)= 3. 17. (5 + 21): 2  = 3.13.17 = 663

c: 1, 13, 17, 663;

d=17 + 21 +....+ 41 ;

d = (17 +41) + (21 + 37) + (25 + 33) + 29 = 58.3 + 29 = 203;

d : 1, 7, 29, 203.