secondary math

30 = 2 x 3 x 5
45 = 3 x 3 x 5
LCM = 180 = 2 x 2 x 3 x 3 x 5
HCF = 5
(1)
Since LCM contains only 2, 3 and 5. No other numbers can exist in x, otherwise LCM will not be 180.
(2)
Since HCF is 5, x must contain a 5 ( and one 5 only because 180 has one 5 only). Also, x must NOT contain 3, otherwise HCF will not be 5.
(3)
So the remaining number in x must be 2. Since 30 have one 2, 45 have no 2 and the LCM 180 have two 2s, so x must contain two 2s.
Therefore x = 5 x 2 x 2 = 20.

Consider HCF(30,45)=15 and LCM(30,45)=90.

Now, you required that HCF(x,15)=5 and LCM(x,90)=180.

Note that:
15 = 3*5
Therefore, x should have a factor 5, but no factor of 3

180 = 2*2*3*3*5
90 = 2*3*3*5

x should have a factor of 2*2.
x should not have a factor of 3 as above mentioned.
x should have a factor of 5, but only one 5.
x should not have other prime factor from the LCM of 180.

For x=2*2*5=20, both conditions are satisfied.

HCF is 5 , so x contains a factor (5)

as, HCF of 30 and 45 is 15, that means, x cannot be divisible by 3.

also,

180=2^2*3^2*5
30=2^3^5
45=3^2*5

because their LCM is 180, so x must have the factors (2^2)
so x = 2^2 * 5 =20

2013-08-12 23:18:56 補充：
30=2*3*5