TITLE: Maps from mod-$p$ Eilenberg-MacLane spectrum to connective $K$-theory spectrum
QUESTION [10 upvotes]: Let $ku$ be the connective cover of the complex $K$-theory spectrum $KU$. Consider the mod-$p$ Eilenberg-MacLane spectrum $H\mathbb{Z}/p$.
I want to see that $[H\mathbb{Z}/p,ku]=0$. 
Since $H\mathbb{Z}/p$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories") the result will follow once we know that $ku$ is harmonic (Corollary 4.11 in loc. cit.)

Is it true that the spectrum $ku$ is harmonic? 

If not, how can I prove the claim? 
Certainly, an idea is to use directly the theorem of Margolis (“Eilenberg-MacLane Spectra”): For any spectrum $Y$ of finite type, there is an isomorphism
$$[H\mathbb{Z}/p,Y]\to \text{Hom}_{\mathcal{A}}(H^{*}(Y,\mathbb{Z}/p),\mathcal{A})$$
where $\mathcal{A}$ is the Steenrod algebra. 

Is it true that $\text{Hom}_{\mathcal{A}}(H^{*}(ku,\mathbb{Z}/p),\mathcal{A})=0$? 

If the answer is yes, why is that?

REPLY [15 votes]: You can also cheat: since HZ/p is connective,$$ [HZ/p,ku] = [HZ/p,KU] = [HZ/p \wedge KU,KU]_{KU-modules} = 0$$ since $HZ/p \wedge KU = 0$.
Edit: Let me stress that the other answers give more information, namely the calculation of maps from HZ/p to ku in all degrees.