Let us take a simple example where $n=1$. When a single digit number, say $a$ is multiplied by $10^1$, the answer comes out to be "$a0$". If a number is $a_1a_2a_3\dots a_k$, then multiplication with $10^1$ gives "$a_1a_2a_3\dots a_k0$".

If number is multiplied with $10^2=100$, then there are $2$ zeroes at the end. Following the same pattern, if a certain number with any decimal part is multiplied with $10^n$, there are $n$ zeroes at the last.

This all is super$^\infty$ easy and obvious, but is there an even simpler or somewhat intuitive reason why these all zeroes are just "attached" to the number, an explanation which is suitable for a small kid who is a newbie at Mathematics?

As far as I think, there is no suitable explanation which is "simple" enough for a kid to understand. Excpeting the "suitable for newbie" clause, is there some interesting interpretation of multiplication with powers of $10$ ?

If we interpret Multiplication as a "Repetitive Addition", for example $a \times b$ is just $\underbrace {a+a+\dots +a}_{\text{b times}}$, then also it is not quite "intuitive" why there suddenly appears a zero out of nowhere when $b=10^c \,\,;c \in \mathbb{N}$.

Hopefully this elementary question isn't too "off-topic" for this website. I think something constructive will appear as a result of discussion over this question.

Thanks in Advance !

  • $\begingroup$ I thinks it's not easy to explain to the kid in terms of interpretations of multiplication, since these attachments of zeros when multiplying by powers of ten are just a property of decimal number system. It's the same for multiplication by e.g. $16^n$ in hexadecimal or $8^n$ in octal systems. $\endgroup$ – Ruslan Mar 3 '17 at 11:48
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    $\begingroup$ There's a nice answer here. You might get others at matheducators.stackexchange.com $\endgroup$ – Ethan Bolker Mar 3 '17 at 19:22
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    $\begingroup$ *isn't too off topic $\endgroup$ – Kunal Pawar May 7 '17 at 10:23

These kinds of patterns occur because of the base in the number system being used. For example, in binary multiplication by 2 is simply adding a zero to the end of the bit string. In fact, in some programming languages, you can perform bit shifts to multiply or divide by a power of 2 since that is faster than performing multiplication or division.

In base 10 system, powers of 10 add zeros to the end of a multiplicand. This is simply a product of using 10 as the base of our number system. Recognizing the result as you have expressed in your question is actually a shortcut to performing multiplication, and understanding how it works is the basis for using scientific notation.

As far as teaching it to a child, well-reasoned programs that teach math and prepare kids for higher math (in the United States, this is called the Common Core State Standards, or CCSS) show these shortcuts to students and have them practice them, in order to strengthen their understanding of the base ten system. For example, in 3rd grade, the CCSS wants students to

Multiply one-digit whole numbers by multiples of 10 in the range 10–90 (e.g., $9\times80$, $5\times60$) using strategies based on place value and properties of operations. (CCSS 3.NBT.3)

At first, students will do the multiplication longhand, but soon they should see a pattern. And since they know the product of $9\times8$, for example, they will soon see that $9\times80=9\times8\times10=72\times10=720$ and be able to do it in one step (that is what the standard means by "strategies based on place value and properties of operations" -- here the property is associativity).

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    $\begingroup$ +1 Nice explanation. The connection with place value is the central idea. Your answer could be stress that even more. It's what teachers in elementary grades do when teaching the "rule". In later grades "append a $0$" becomes "move the decimal point (right or left)". $\endgroup$ – Ethan Bolker Mar 3 '17 at 19:21

As far as I think, there is no suitable explanation which is "simple" enough for a kid to understand.

The "multiply by $10$ rule" is an artifact of the base-$10$ (decimal) representation, and is a direct consequence of the fact that the next number after $9$ is $10\,$. Kids are taught early that counting goes:

$$ 1\,, \;2\,, \;3\,, \;4\,, \;5\,, \;6\,, \;7\,, \;8\,, \;9\,, \;\color{red}{10} $$

This makes it intuitively obvious that adding $1$ ten times equals $10\,$:

$$ \underbrace{1\, + \,1\, + \,1\, + \,1\, + \,1\, + \,1\, + \,1\, + \,1\, + \,1\, + \, 1}_{\text{10 times}} \, = \, 10 $$

Now consider that $a = \underbrace{1 + 1 + \cdots + 1}_{\text{a times}}\,$. Then multiplying $a$ by $10$ can be thought of as:

$$ \begin{align} & \underbrace{\underbrace{1 + 1 + \cdots + 1}_{\text{a times}} \;+\; \underbrace{1 + 1 + \cdots + 1}_{\text{a times}} + \cdots \underbrace{1 + 1 + \cdots + 1}_{\text{a times}}}_{\text{10 times}} \\[5px] =\; & \underbrace{\underbrace{1 + 1 + \cdots + 1}_{\text{10 times}} \;+\; \underbrace{1 + 1 + \cdots + 1}_{\text{10 times}} + \cdots \underbrace{1 + 1 + \cdots + 1}_{\text{10 times}}}_{\text{a times}} \\[5px] =\; & \underbrace{\;10 \;+\; 10 \;+\; \cdots \;+\; 10}_{\text{a times}} \\[5px] \end{align} $$

The latter form makes it somewhat more obvious that an addition similar to $a = \underbrace{1 + 1 + \cdots + 1}_{\text{a times}}\,$ gives $10\,a = \underbrace{10 + 10 + \cdots + 10}_{\text{a times}}\,$, and since the RHS has an extra $0$ at the end of each term, the LHS will be the original number with an extra $0$ at the end.

Once the multiplication by $10$ is understood and accepted, the rule for $10^n$ becomes obvious by repetition (more formally, by induction).


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