The lowest number that is a multiple of both 60 and the integer n is 180. Find the smallest possible value of n. I have one solution but I think it's just a wild guessed one. Tell me if I am correct and also if not, then how should it be done?
What I have done is divided 180 by 60 to get 3. Then take lcm of 60 and multiples of 3 one by one until I get 180 as the lcm.
So I would first take the lcm of 60 and 3, then of 60 and 6, then of 60 and 9 after which I would get 180.
Is this the correct way to solve it?
P.S This is just a 1 mark question of one of the paper of O Levels in which calculators are not allowed and not more than 1 or 2 blank lines are given for working :/. Keep that in mind. :P
 A: $$\operatorname{lcm}(60,n) = 180$$
$$\operatorname{lcm}(2^2 \cdot 3^1 \cdot 5^1,
                    2^{n_2} \cdot 3^{n_3} \cdot 5^{n_5})
 = 2^2 \cdot 3^2 \cdot 5^1$$
$$ 2^{\max(2,n_2)} \cdot 3^{\max(1,n_3)} \cdot 5^{\max(1,n_5)}
   = 2^2 \cdot 3^2 \cdot 5^1$$
\begin{align}
    \max(2,n_2) &= 2\\
    \max(1,n_3) &= 2 \\
    \max(1,n_5) &= 1
\end{align}
The smallest possible exponents are thus
\begin{align}
    n_2 &= 0\\
    n_3 &= 2 \\
    n_5 &= 0
\end{align}
So $ n = 2^0 \cdot 3^2 \cdot 5^0 = 9$.
A: $n$ is the smallest number that divides 180 but not 60. (If it divided both, 180 couldn't be the least common multiple because $60 < 180$.)
The primes in $60$ are $(2, 2, 3, 5)$ and the ones in $180$ are $(2, 2, 3, 3, 5)$.
If $n$ divides $180$, its primes are contained in  $(2, 2, 3, 3, 5)$.
The smallest list contained in $(2, 2, 3, 3, 5)$ but not in $(2, 2, 3, 5)$ is $(3, 3)$, and $3 \cdot 3 = 9$.
A: Given two positive integers $A,B$, you have $AB=gcd(A,B) lcm(A,B)$.
Here, $A=n, B=60$ and $lcm(A,B)=180$. Hence, calling $g$ the gcd of $n$ and $60$: $60n=180g$, or $n=3g$.
Thus $n$ is a multiple of $3$. But then, the gcd is also a multiple of $3$, since $3$ divides $60$. Hence $n$ is a multiple of $9$. The smallest multiple of $9$ is $9$ itself, and you can check it works.
A: Totally without the use of any mathematics: 
x = 1, 2, 3, 4, 5, 6 are already divisors of 60, so the lowest common multiple of 60 and x is 60. 
180 is not a multiple of x = 7 or 8, therefore it is not the lowest common multiple of 60 and x; it is not a multiple of x at all. 
Let x = 9: Yes, the lowest common multiple of 60 and 9 is 180. Because the multiples of 60 are 60, 120, 180, 240, ... and the first of those that is also a multiple of 9 is 180. 
(In the general case, factoring the numbers that correspond to 60 and 180 into prime factors will give you the solution quickly). 
A: Even though there are lots of correct answers (even one accepted) I will add another. The OP asks

Is this the correct way to solve it?

and adds the important information

P.S This is just a 1 mark question of one of the paper of O Levels in
  which calculators are not allowed and not more than 1 or 2 blank lines
  are given for working :/. Keep that in mind. :P

I think that in this case his method is just the right one to use. It requires minimum use of fancy stuff and takes minimal time when there's time pressure. It may well be what the examiners had in mind when they asked the question. 
A: You can do a prime factorization of each number:
$$60 = 2^2 \cdot 3^1 \cdot 5^1$$
$$180 = 2^2 \cdot 3^2 \cdot 5^1$$
The number $180$ is the least common multiple of $60$ and this other number.  As the LCM, it takes on the largest exponent for each prime of any of the numbers.
The exponents on the $2$ and $5$ are the same in the LCM as for $60$, so the other number need not have any factors of $2$ or $5$.
But we do need two factors of $3$.
Hence ... ?
A: $$180=2^2.3^2.5$$
$$60=2^2.3.5$$
so $n$ must have the same primes as $180$
$$n=2^i3^j5^k$$ where $i,j\leq 2$ and $k\leq 1$
$60$ is not a multiple of $n$, otherwise their $lcm$ would have been $60$.
 Therefore, we must have $$n\equiv 0 \pmod {3^2}$$ It follows the smallest $n$:  $$n=9$$
A: $$180=2^2.3^2.5$$
$$60=2^2.3.5$$
so $n$ must have a similar prime factorization as $180$
$$n=2^i3^j5^k$$ where $i,j\leq 2$ and $k\leq 1$
$60$ is not a multiple of $n$, otherwise their $lcm$ would have been $60$.
 Therefore, $n$ must be a multiple of $9$. That is only to ensure that $n$ does not divide $60$. It follows the smallest $n$:  $$n=9$$
A: It seems to me (that is, I offer no mathematical arguments to support my proposal) that $-180$ is the smallest possible integer for which $180$ is a multiple. 
The question as posed does not seem to me to strictly insist or restrict that the answer be a positive integer; in fact, it seems to specifically allow for such possibility by avoiding usage of nearly all mathematical jargon except for "integer" (which may be negative or positive) and "multiple" (again, may be negative or positive). 
Also $-180$ is clearly smaller than $9$, unless "smallest possible value of $n$" does not mean what it would appear to mean to most mortals familiar with a number line. 
A: No need to big calculations. Simple justification is required.
The number n should be factor of 180 but should not be factor of 60. 
The factors of 60 are 1,2,3,4,5,6,10,15,20,30 & 60. 
The factors of 180 are 1,2,3,4,5,6,9,10,12,15,18,20,30,36,45,60,90 & 180. 
It is clear that the lowest possible number which is a factor of 180 but not that of 60 is only 9
Hence answer is 9
