Have you ever wanted to know how many zero’s were at the end of a number?
Probably not very often, unless you compete in math competitions or just like subjecting yourself to puzzling math questions.
Are you wondering what a Terminal Zero is? They are just 0’s at the end of a number, like the 0’s highlighted here in red.
$${123,\color{red}{000}}$$
The recent MOEMS® Problem of the Week posted on April 29th, 2024 asked the following question:
In the number 203,500, the last two zeros are called the terminal zeros.
If 30 x 40 x 50 x 60 x 70 is done, how many terminal zeros would the product have?
MOEMS® Problem of the Week
Can you answer this problem? How long does it take you to answer it? Click the expander at the left to see the answer.
Answer: 30 x 40 x 50 x 60 x 70 = 252,000,000
There are 6 terminal zeros.
MOEMS® Problem of the Week
The problem is easily solved without a calculator. What techniques might we try?
We could multiply all the numbers and look at the result. That seems time consuming and error prone without a calculator.
All of the numbers in the example are multiples of 10. We could change the problem to \((10 \cdot 3) \cdot (10 \cdot 4) \cdot (10 \cdot 5) \cdot (10 \cdot 6) \cdot (10 \cdot 7)\), most everybody would probably guess there are at least 5 terminal 0’s.
$$100,000$$
The question remains, how many more terminal zeros exist in \((3 \cdot 4 \cdot 5 \cdot 6 \cdot 7) \cdot 100,000\).
We could multiply them all to find out:
$$ \begin{align} 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7&=252\color{red}{0}\\ 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7\cdot 100000&=252\color{red}{0} \cdot 1\color{red}{00,000} \\ &=252,\color{red}{000,000} \end{align} $$
Here we see that our product has 6 terminal zeros.
Is there a faster way to figure this out? We know that we will have 5 terminal zero’s already, so let’s put that part of the problem aside.
Let’s isolate our problem to determining how many terminal zero’s are found in the product of $$3\cdot4\cdot5\cdot6\cdot7$$.
What do we know for sure:
Any number with \(10\) as a factor will end with a \(0\). Can we make a \(10\)?
To make a \(10\), we need at least one \(2\), and at least one \(5\). We have a \(5\), and we have the factor \(2\) in both \(4\) and \(6\).
$$ \begin{align} 10 = 5^1*2^1 \\ 100 = 5^2*2^2 \\ 1000 = 5^3*2^3 \\ \end{align} $$A \(10\) exists in our number, so we know we have one terminal zero, do we have a second terminal \(0\)?
To figure this out we’d need to find a multiple of \(100\). To make a multiple of \(100\) we need a \(2\) and a \(5^2\). The prime factorization of the product of \(3\cdot4\cdot5\cdot6\cdot7\) is \(2^3*3^2*5^1*7^1\)
The product contains a \(2^2\) but it does not contain a \(5^2\) so it is not a multiple of \(100\), and therefore only contains the single terminal \(0\) found as a result of being a multiple of \(10\).
Adding our one terminal zero to the previous five, makes 6 terminal zeros.
What is the smallest number we could multiply the product by to add another terminal 0?
Since we already have \(2^2\) and \(5^1\) we only need to multiply by 5, or any multiple of 5 and our product will become a multiple of 100. You can try multiplying our product by any of the following 5, 10, 15, 20 and you’ll see that exactly 1 terminal zero is added, until you get to 25.
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