Numbers $n$ whose prime factors are $2$ and $5$ if and only if $\sum_{k=1}^{rad(n)}\mu(k)k=0$?

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In the literature if defined the arithmetical function $rad(n)$ as $1$ if $n=1$ and for $n>1$ by $rad(n)=\prod_{p|n}p$. It is obviously a multiplicative function. Too we know the Mobius function $\mu(n)$ debined by $\mu(1)=1$, $\mu(n)=(-1)^k$ if $n$ is a product of $k$ distinct primes, and is equals to $0$ otherwise.

Two hours ago I had my head in other problem when I've write this nice, really I think that is a nice statement perhaps since our numeric system is based in powers of $2$ and $5$. I believe that isn't obvious. I don't know if this exercise is in the literature.

Question. Prove or refute this statement for integers $n\geq 1$: $n$ is a term of the Sloane sequence A033846, numbers n whose prime factors are 2 and 5, if and only if $$\sum_{k=1}^{rad(n)}\mu(k)\cdot k=0.$$

Example. Let $n=10$ the first term of cited sequence, then

$$\sum_{k=1}^{10}\mu(k)k=1\cdot1-1\cdot2-1\cdot3+0 \cdot4-1\cdot5+11\cdot6-1\cdot7+0\cdot 8+0\cdot 9+1\cdot 10=0$$

I excuse put this question here since I believe that at least one of the implications could be easy for you. Or a counterexample if the conjecture is false. Can your give the details and ensure a nice answer? Thanks in advance.

$$\color{red}{\text{Update: There was a mistake in my genuine question. Now the question is rigth. Thanks.}}$$

References:

[1] The On-Line Encyclopedia of Integer Sequences, A033846 Numbers n whose prime factors are 2 and 5. https://oeis.org/A033846