$\widetilde{H}^n(X,\Bbb Z_p)=0$ for all $n$, odd primes $p$ implies $\widetilde{ H}^n(X,\Bbb Z)$ consists entirely of elements of order a power of $2$ Let $X$ be a CW complex with finitely generated homology groups. We know that from the universal coefficient theorem(UCT) that $$H^n(X;G)\cong \textrm{Ext}(H_{n-1}(X),G)\oplus \textrm{Hom}(H_n(X),G)$$
for any abelian group $G$.
Suppose $\widetilde{H}^n(X,\Bbb Z_p)=0$ for all $n$ and for all odd primes $p$. Then how does the UCT imply that $\widetilde{ H}^n(X,\Bbb Z)$ consists entirely of elements of order a power of $2$?
(I was reading Hatcher's Example 3E.5.)
 A: The assumption implies in particular that $\hom(H_n(X),\mathbb Z/p) = \mathrm{Ext}(H_n(X),\mathbb Z/p) = 0$ for all $n$ and odd primes $p$. 
Using the long exact sequence associated to $0\to \mathbb Z\to \mathbb Z\to \mathbb Z/p\to 0$, we get that $\hom(H_n(X),\mathbb Z)$ is uniquely $p$-divisible (that is, multiplication by $p$ is an isomorphism), and same thing for $\mathrm{Ext}(H_n(X),\mathbb Z)$. 
Now the first thing implies that $\hom(H_n(X),\mathbb Z) = 0$. Indeed, suppose $f$ is a nonzero morphism, then there are $x,k$ with $f(x) = k\neq 0$. Now if $f$ were $p$-divisible for all odd primes $p$, it would mean that $k$ is too, which is absurd. So $\hom(H_n(X),\mathbb Z) = 0$. 
Since now $H_n(X)$ is finitely generated (you said so in the comments), this is enough to conclude : indeed by the structure theorem, $H_n(X) \cong \mathbb Z^r \oplus \bigoplus_p\bigoplus_i \mathbb Z/p^{\alpha_i}$ for some $r$ and $\alpha_i$'s, so that this equality to $0$ implies $r=0$, and the condition on the $\mathrm{Ext}$'s implies that all odd components are $0$. 
So $H_n(X)$ is $2$-torsion, and therefore $H^n(X)\cong \mathrm{Ext}(H_n(X),\mathbb Z)$ is too
I'm not entirely sure what happens if you don't assume $H_n(X)$ to be finitely generated, but my guess is that the result doesn't hold. 
