How many of the 0 digits are place holders in 330.606?

Question

How many of the 0 digits are place holders in 330.606?

This is the problem. I have tried solving it on the online calculator.

Answer 1

I see two such zeros serving as place holders: one $0$ for the "one's" place, the other for the one-hundredth's $\left(\frac 1{100}\text{th's}\right)$ place.

You have, specifically $$3 \times 100 + 3\times 10 + \color{blue}{\bf 0} \times 1 + 6 \times \dfrac 1{10} + \color{blue}{\bf 0}\times \dfrac 1{100} + 6 \times \dfrac 1{1000}$$

Neither zero can be removed without changing the position of the other numbers: which would then result in a number that is not at all equivalent to the given number.

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In contrast: we can see that $\color{red}{\bf 0}.3\color{blue}{\bf 0}4\color{red}{\bf 0}$ has $\color{red}{\bf two}$ removable zeros, and $\color{blue}{\bf one}$ nonremovable zero which serves as a place holder.

Answer 2

As a general rule, if you can remove a $0$ digit from a number and keep the same value, it is a leading $0$. In any other case, it's a placeholder.

Let's consider your number $330.606$

Both zeroes here are placeholders because they actually contribute something to the number. You can't remove them and sustain the number's value. (Of course, if you're talking about physical quantities, there is a difference between saying 1.00 grams and 1 gram, but that's unrelated to this problem ;) )