If $G = C_{25}\times C_{45}\times C_{48}\times C_{150}$, where $C_n$ denotes a cyclic group of order $n$, how many elements of order 5 does $G$ have? Suppose that $G = C_{25}\times C_{45}\times C_{48}\times C_{150}$, where $C_n$ denotes a cyclic group of order $n$.
How many elements of order 5 does $G$ have?
I realize since each $C_n$ is cyclic, we can write $C_n = <c>$ with $c^n = e$. Hence, in $C_{25}$ for example, $C_{25} = <c> = \{c,c^2,c^3,\cdots,c^{25} = e\}$. The elements with order 5 are precisely $c^5, c^{10},c^{15}$, and $c^{20}$. Similarly, we find 4 elements in $C_{45}$ and $C_{150}$ of order 5.
Now, an element $<a_1,a_2,a_3,a_4>$ of order 5 can have one $a_i$ of order 5 $(3*4$ ways to do this), 2 $a_i$ of order 5 $(4^2*3$ ways to do this), or 3 $a_i$ of order 5 $(4^3$ ways to do this), with all other $a_i = e$.
Hence, I think my answer is $12+4^2*6+4^3$. Does this seem right? Is there a better way to do this?
 A: Let $g \in C_{25}\times C_{45}\times  C_{48}\times  C_{150} = g_{25}\times g_{45}\times  g_{48}\times  g_{150}$.
$g$ is of order $5$ if and only if 
$$  
(g_{25}^5 = e_{25}) \wedge (g_{45}^5 = e_{45}) \wedge (g_{48}^5 = e_{48}) \wedge (g_{150}^5 = e_{150}) (\wedge g \neq e)
$$
That last $g \neq e$ is there because if all four projections of $g$ are the identities in their respective cyclic groups, then $g$ is order $1$ rather than $5$.
Now in any  $C_{5K}$ there are exactly  $5$ elements $f$ such that $f^5 = e$; one of those is, of course, the identity element itself.  And in any $C_n$ where $n$ is not a multiple of $5$, only the identity has $f^5 = e$.
So to form an element of $G$ that is of order $5$ one must choose one out of five possibilities in three projections, and the identity in the $C_{48}$ projection, but avoid choosing the identities in every projection.
Thus the total number of elements of $G$ that are of order $5$ is
$$
5\cdot 5\cdot 5 -1 = 124$$
Your reasoning, with Steve D's correction, gives the right answer.
