Question

si:MCD(2A;2B)=18
HALLAR:MCD(9A,9B)

Answer 1

Si: A=αk ; B=βk , donde α,β son primos entre sí (MCD=1) entonces:

MCD(2A;2B)=MCD(2αk;2βk)=2k*MCD(α;β)=18
 2k*1 = 18
  k = 18/2
  k = 9

Si k=9:

MCD(9A,9B)=9k*MCD(α,β) = 9k*1 = 9(9)

MCD(9A,9B) = 81

Saludos!