Question

suma a patru numere naturale impare consecutive este 1320.

Answer 1

a + (a + 2) + (a + 4) + (a + 6) = 1320
4a + 12 = 1320
4a = 1320 - 12 = 1308
a = 1308:4 = 327
Numerele sunt: 327, 329, 331, 333

Answer 2

1nr=x
2nr=x+2
3nr=x+4
4nr=x+6
x+x+2+x+4+x+6=1320
4x+12=1320
4x=1308
x=1308:4
x=327
x+2=329
x+4=331
x+6=333