If x is a positive integer whose unit digit is 0, is x divisible by 6?
(1) 3x has fewer factors than does 2x.
(2) 6x has the same number of factors as does 5x.
x=(2^a)*(3^b)*(5^c)*(7^d)**(p^z)
a'>'=1, c'>'=1 and b,d....z'>'=0.
number of factors for x=(a+1)*(b+1)*(c+1)*(d+1)***(z+1)
Is b'>'0 ?
Statement 1:
number of factors for 3x=(a+1)*(b+2)*Y; where Y=(c+1)*(d+1)***(z+1)
number of factors for 2x=(a+2)*(b+1)*Y
(a+1)*(b+2)*Y'<'(a+2)*(b+1)*Y
b'>'a
Sufficient.
Statement 2:
number of factors for 6x=(a+2)*(b+2)*(c+1)*Z; where Z=(d+1)***(z+1)
number of factors for 5x=(a+1)*(b+1)*(c+2)*Z.
(a+2)*(b+2)*(c+1)*Z=(a+1)*(b+1)*(c+2)*Z.
if b=0, than 2*(a+2)*(c+1)=(a+1)*(c+2)
this works out to:
ac+3c+2=0 which is not possible since a,c'>'0.
therefore b'>'0.
Sufficient.
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