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I tried uploading a gif, but was unable to do so. What I can do, is share a link to the gif here. (SE software seems to have allowed me to share the link, but not upload it.)

What it shows, initially, is the times table for 9, without answers:

$\begin{matrix} 1\times 9 = \\2\times 9 = \\ 3\times 9 = \\ 4\times 9 = \\ 5\times 9 = \\ 6\times 9 = \\ 7 \times 9 = \\ 8\times 9 = \\ 9\times 9 = \end{matrix}$

Then the student proceeds by starting from $2\times 9$ with the initial digit 1, then counts up from one as she moves down this column in the times table, to get:

$\begin{matrix} 1\times 9 = \\2\times 9 = 1\\ 3\times 9 = 2 \\ 4\times 9 = 3\\ 5\times 9 = 4\\ 6\times 9 = 5 \\ 7 \times 9 = 6\\ 8\times 9 = 7\\ 9\times 9 = 8\end{matrix}$

Then they show the student counting backwards (starting from 1 at the bottom of the list, up to 9 at the top of the list, in the unit digit's place, to get:

$\begin{matrix} 1\times 9 = \; 9\\2\times 9 = 18\\ 3\times 9 = 27 \\ 4\times 9 = 36\\ 5\times 9 = 45\\ 6\times 9 = 54 \\ 7 \times 9 = 63\\ 8\times 9 = 72\\ 9\times 9 = 81\end{matrix}$

I've never seen this before. Is this primarily a "rote computational trick" for elementary students to learn single-digit multiplication by $9$, or is there an explanation that elementary students can understand as to "why this works"? I have done graduate research in secondary ed, and undergraduate math ed, and teach at a University, so I am not well versed wrt math education at the primary ed level.

I'm not sure what I think about this "trick"; I would hope at least, for primary grade students, that subsequently, why this works for 9, would be explained to students. It isn't entirely symmetrical, either, as the student proceeds, with the ten's digit from the top $1\times 9$ being blank (or zero), then descending by adding 1 down to 8, to complete the ten's digit. Then she starts from the last row to enumerate unit digits, moves upward each step, counting from 1 to 9.

ALSO, because it doesn't seem to be clear to all answerers, I am firstly interested in how to help primary students learn the multiplication tables. There are clever tricks, but often times students cannot understand WHY they work, until some future time. I'm reluctant to demand rote memorization, so if there are reasonable tricks, accompanied with explanations that make sense to students first learning their multiplication tables, I would like to know those explanations.

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    $\begingroup$ Funny, I have no memory of anyone ever teaching me this, but this is exactly how I have had multiplication by 9 up to 99 memorized since I was around 6 or 7 (I'm in my mid 40s now). I noticed back then that as you go up by 9, you are actually just incrementing the first digit by 1, and then incrementing the second digit down from 9 to 0. I always got a kick out of being able to count to 99 by 9 in a few seconds. But then I always did math "my own way" and the ways I was taught in school never made sense to me and seemed dreadfully inefficient. $\endgroup$ – Jonathan van Clute Dec 14 '19 at 23:18
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    $\begingroup$ Note that this works for any base n where you're looking at the multiplication tables for n-1 (i.e. the highest single digit). Binary (n-1 = 1): 01, 10 Ternary (n-1 = 2): 02, 11, 20 Octal (n-1 = 7): 07, 16, 25, ..., 61, 70 Hexadecimal (n-1 = F): 0F, 1E, 2D, 3C, ..., E1, F0 and so on... $\endgroup$ – Flater Dec 16 '19 at 9:11
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    $\begingroup$ This was the subject of a song on Square One: youtube.com/watch?v=8Gna9nME5DU $\endgroup$ – user560822 Dec 16 '19 at 16:26
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    $\begingroup$ The same trick also works for $99$. Starting with $1 \to 0099, 2 \to 0198, 3 \to 0297 \dots$ and ending with $97 \to 9603, 98 \to 9702, 99 \to 9801$ $\endgroup$ – Steven Gregory Dec 16 '19 at 16:51
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    $\begingroup$ I learned this as “the number you're multiplying, minus one; and then the number you need to make them total nine” and that's how I do it still. (Maybe easier to think of as “multiplicand minus one; nine minus that.”) The difference being that it works for an individual number, independent of the list. $\endgroup$ – Jacktose Dec 16 '19 at 22:50
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Note $9 = 10-1$ so:

$$ 5 \times 9 = 5 \times (10-1) = 50 - 5 = 45, $$ and the same for all the others: $$ 8 \times 9 = 8 \times (10-1) = 80 - 8 = 72. $$ This works for $k \times 9$ where $1 \le k\le 10$.
Although we always have $$ k \times 9 = (k-1)\times 10 + (10-k) , $$ this is the final decimal answer only when $1 \le k \le 10$.


After the kids do this, ask them to try to come up with a similar rule for multiplication by $99$. (That one will work for $1 \le k \le 100$.)

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  • $\begingroup$ Multiplication by $11$ is also quite straight forward, particularly multiplying $1, 2, 3, 4, 5, 6, 7, 8, 9$ by $11$, when introducing double digits: $\begin{matrix} 1\times 11 = 11 \\2\times 11 = 22 \\ 3\times 11 = 33\\ 4\times 11 = 44\\ 5\times 11 = 55 \\ 6\times 11 = 66\\ 7 \times 11 = 77\\ 8\times 11 = 88\\ 9\times 11 = 99\end{matrix}$ $\endgroup$ – Namaste Dec 14 '19 at 15:05
  • $\begingroup$ For the explanation you provide, students must already understand the distributive rule of multiplication over a sum (or difference). $a(b +(-c)) = ab + a(-c) = ab - ac$. I think primary students first learning their multiplication tables (say single digit factors), have not yet been exposed to the distributive law of a factor over a sum. $\endgroup$ – Namaste Dec 14 '19 at 15:11
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    $\begingroup$ @Namaste: Agreed. So this answer is more for the teacher than for the student. If the teacher is afraid of algebra, then when I put a $k$ in there, I lose even them. $\endgroup$ – Gerald Edgar Dec 14 '19 at 15:17
  • $\begingroup$ Indeed. I like it for that reason! $\endgroup$ – Namaste Dec 14 '19 at 22:39
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Anything that is just a trick leads to students having wrong ideas about what math is. But methods that help students see the patterns can help them learn the multiplication facts, along with getting a better feel for what's going on.

I'd call this a way to think about 9s. (There are many.) This method shows that you add 10 for each new nine, and then take away 1.

Multiplying by 4 is doubling and doubling again. Multiplying by 5, they should notice that odd numbers times 5 result in a unit digit of 5 and even numbers time 5 result in a unit digit of 0. Exploring patterns while learning these results is vital.

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    $\begingroup$ Thanks, Sue. I concur with you. I've been thinking more about this, and was going to comment: "I wonder whether in primary grades, introducing multiplication for prime numbers: 2, 3, 5, 7 might serve students well, when expanding onward to learn the complete tables for 1-9. I mean, e.g. $4 = 2\times 2$, $6 = 2\times 3$, $8 = 2\times 4$, $9=3\times 3$," but held off. I think the "doubling" you speak of makes $2$'s time tables all the more salient. $\endgroup$ – Namaste Dec 13 '19 at 18:43
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    $\begingroup$ I mean, e.g., if the times table for $2$ has been covered and mastered, then, $2\times 2\times 2$ can be solved by knowing $2 \times 2= 4$, and that $2\times 4 = 8$. But this might require their knowledge that $2\times 4 = 4\times 2$. Maybe this idea is better understood later! $\endgroup$ – Namaste Dec 13 '19 at 18:58
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    $\begingroup$ It is so cool to see kids learning about multiplicative commutativity. To us adults, it seems so obvious, but it's not obvious to young kids. But I don't think that one (2x2x2) needs commutativity. $\endgroup$ – Sue VanHattum Dec 14 '19 at 0:08
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    $\begingroup$ By order of operations, calculating $2\times 2\times 2$ would lead to the intermediate step, $(2\times 2)\times 2 = 4\times 2=8$. But I don't think order of operations for all arithmetic operators would have been covered at this stage, unless a teacher explains proceeding from left to right when there is only one arithmetical operation. Hence I thought of the commutative for multiplication, to help explain to students why some came up with $2\times 2\times 2 = 4\times 2 = 8 = 2\times 4 = 2\times 2 \times 2.$ $\endgroup$ – Namaste Dec 14 '19 at 15:27
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    $\begingroup$ @Namaste That would be better explained with the associative property IMO, as $(2*2)*2=2*(2*2)=4*2=2*4=8$ $\endgroup$ – Abion47 Dec 14 '19 at 17:33
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Yes. This is also a trick that you can do on your fingers, too. For instance, let's say you wanted to calculate $9\times3$.

enter image description here

Hold out your hands and bend your third finger down as shown. So nine fingers are "up" (fingers up, $9$, finger #3 down. (9x3). You have two fingers to the left of the bent finger and seven to the right, indicating the product of $27$.

Explaining this to students will depend on their prior knowledge and your standards. I would probably note that $9=10-1$, and so adding nine to a number is the same as adding one to the ten's place and subtracting one from the unit's place (as long as there was something to subtract in the unit's place).

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    $\begingroup$ @Namaste I agree. I saw a YouTube video a while ago (that I don't think I'd be able to find again) that argued that 2, 3, 4, and 5 are fairly easy to memorize by step-count, and 9, 10(, and 11) are fairly easy to memorize by tricks. So that just leaves six multiplication facts ($6^2, 7^2, 8^2, 6\times7, 6\times8, 7\times8$) that need to be memorized just by committing them to memory. So I guess it's a case that every little bit helps! $\endgroup$ – Matthew Daly Dec 13 '19 at 19:01
  • $\begingroup$ So you are arguing that teaching tricks (which need to be memorized) and memorizing a few multiplications, is all that primary students need? You did not answer my question with respect to whether procedural/rote/memorized is a good idea, or if explanations to students to understand the "why" need to be included. And, we are talking about multiplication, and not adding numbers. $\endgroup$ – Namaste Dec 15 '19 at 15:32
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    $\begingroup$ I mainly work with adult students. Some of them love this trick, and for adults who haven't gotten their times facts, I figure love is good enough. But I have never liked it, because it hides the reasoning.(I'm guessing this would still work (for 7 instead of 9) if we had 8 fingers and used base 8. $\endgroup$ – Sue VanHattum Dec 16 '19 at 15:19
  • $\begingroup$ I am currently a 27 year old software engineer, and I still do this. I don't usually do the full gesture, but I'll look at my hands and look at the 3rd finger and visualize it. $\endgroup$ – Cruncher Dec 16 '19 at 17:36
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    $\begingroup$ Being taught this when I was struggling learning to multiply quickly was the first time someone showed me that multiplication worked in a pattern. It was always just described to me as "repeated addition", so that's what I did until then. My math teacher was trying to get me registered as "special needs" until then, because I was having so much trouble - by the end of that year, she'd signed off on my "gifted" program instead. $\endgroup$ – gatherer818 Dec 16 '19 at 21:13
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To add on to the other answers, the reason this works is that we use the decimal system, a.k.a. the base-10 system, for our everyday maths. The multiples of the number that is one less than the base results in a phenomenon where the second digit increases at the same rate as the first digit decreases.

$$ 9 * 1 = 09\\ 9 * 2 = 18\\ 9 * 3 = 27\\ 9 * 4 = 36\\ 9 * 5 = 45\\ etc... $$

As the above implies, you see this with different multiples in different base-$n$ systems. For example, the multiples of 7 in octal (base-8):

$$ 7 * 1 = 07\\ 7 * 2 = 16\\ 7 * 3 = 25\\ 7 * 4 = 34\\ 7 * 5 = 43\\ etc... $$

... and $F$ in hexadecimal (base-16):

$$ F * 1 = 0F\\ F * 2 = 1E\\ F * 3 = 2D\\ F * 4 = 3C\\ F * 5 = 4B\\ F * 5 = 5A\\ F * 6 = 69\\ etc... $$

Technically binary (base-2) does it too, though it's less interesting to look at:

$$ 1 * 1 = 01\\ 1 * 2 = 10 $$

(I cheated a little there to keep it formatted nicely as 2 doesn't exist in the binary system, but hopefully the point is still conveyed.)

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    $\begingroup$ +1 for the octal reference. It was always my excuse for confusing Halloween with Christmas, because..... OCT31 = DEC25 $\endgroup$ – JTP - Apologise to Monica Dec 15 '19 at 12:02
  • $\begingroup$ F * 6.. Nice. $\endgroup$ – Cruncher Dec 16 '19 at 17:38
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One neat trick that might help for some of your students is the digital root of a number.

For a number, the sum of its digits is taken. For example,

$$ sumDigits(12345) = 1 + 2 + 3 + 4 + 5 = 15 $$

While the result is not a single-digit number, keep repeating the process:

$$ dr(12345) = 1 + 2 + 3 + 4 + 5 = 15\\ = dr(15) = 1 + 5 = 6 $$

The base case is a single-digit number: its digital root is itself. So $dr(3) = 3$, for example.

A useful trick for remembering the 9x tables specifically is that the digital root of any positive multiple of 9 is always 9!

$$ dr(9) = 9\\ dr(18) = 1 + 8 = 9\\ dr(27) = 2 + 7 = 9\\ dr(36) = 3 + 6 = 9\\ dr(45) = 4 + 5 = 9\\ dr(54) = 5 + 4 = 9\\ dr(63) = 6 + 3 = 9\\ dr(72) = 7 + 2 = 9\\ dr(81) = 8 + 1 = 9\\ ...\\ dr(909) = 9 + 0 + 9 = dr(18) = 1 + 8 = 9 $$

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    $\begingroup$ This is a good observation, but perhaps not yet appropriate for students who are just learning their multiplication facts. $\endgroup$ – Matthew Daly Dec 14 '19 at 16:16
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    $\begingroup$ I think even if you just said that adding up every number in a multiple of 9 makes 9, that's a simple and easily-remembered trick. $\endgroup$ – Daniel Soutar Dec 14 '19 at 18:59
  • $\begingroup$ And a great way to check your answers. $\endgroup$ – Sue VanHattum Dec 16 '19 at 15:20
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In answer to the question "I've never seen this before" I knew this 'trick' with the 9x table and the one with 11x table also mentioned here when I went to primary school about 50 years ago.

I will expand on the 11x as it has already been posted. This is mainly for the teacher but I was about 10 years old when I first learnt it.

With 11 you start with:

1 × 11 = 11
2 × 11 = 22
3 × 11 = 33
.....
When you get to:

10 × 11 = 110
11 × 11 = 121
12 × 11 = 132
13 × 11 = 143
14 × 11 = 154
15 × 11 = 165
16 x 11 = 176

The pattern emerges.

Later you can multiply very large numbers with a 'trick'.

216354 x 11 = 2379894 which you can do in your head in seconds.

You write down the first number (2) then add the next two numbers together and write down the answer, repeat for the following pairs and at the end write down the (4).

2   1     6     3     5     4    
2 2+1=3 1+6=7 6+3=9 3+5=8 5+4=9  4
2     3     7     9     8     9  4

If they add to 10 or more it is not so easy to do in your head as you have to carry the 10.

276354 x 11 = 3039894

2     7     6      3     5     4
2   2+7=9 7+6=13 6+3=9 3+5=8 5+4=9 4
2       9     13     9     8     9 4
3(2+1)  0(9+1) 3     9     8     9 4
3       0      3     9     8     9 4

There are a few other nice 'tricks'.

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    $\begingroup$ I did not post my question to create a "big list" of "tricks when multiplying." Yes, a number of posts have expanded on a trick that explains only 9, or explains only 11, etc. I am interested in knowing whether students, if taught these tricks, also get support in helping them understand why they work. If one must memorize tricks for multiplication of 1 through 12 by 1 through 12, they may as well memorize the table of each of 12 factors, instead of memorizing 12 "tricks" they likely don't understand. $\endgroup$ – Namaste Dec 15 '19 at 0:29
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    $\begingroup$ In this sense, you haven't answered my question. I'm not interested in how clever you might have been, or others. I'd prefer to also hear from primary school teachers who regularly teach the times tables, and what they believe works best, and/or research in primary math education that can inform me/us as to proven practices to successfully help students master their times tables, and whether this subject is taught procedurally, and/or conceptually, or procedurally, and a little while later down the road, conceptually. $\endgroup$ – Namaste Dec 15 '19 at 0:35
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    $\begingroup$ Well, as stated, as a pupil of 10 years old I found this kind if thing fascinating and it helped me develop a love of Math. My Wife is a primary school teacher in the UK and has never seen it used and it is not in the curriculum. I hope this answers your question. $\endgroup$ – Anthony Sach Dec 15 '19 at 20:25
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    $\begingroup$ If you read what I wrote you would realise that I did not say I was first encountering times tables at 10, I was much younger. I did say I was about 10 when I learnt those tricks when I went to primary school. I did not say I learnt them at primary school. It was about 50 years ago, I may have been younger, I do not remember my exact age. I used to get a lot of logic puzzle books and the like. $\endgroup$ – Anthony Sach Dec 16 '19 at 1:05
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    $\begingroup$ Students who are engaged typically share Anthony's enthusiasm for patterns like this, and exploring the pattern can reinforce a student's love of math. There is a way to see if 11 is a factor of a large number that depends on this. To me, the question you asked revolves around how these patterned methods relate to both understanding and getting the times facts into your head. I never use the word trick, because I don't see these as tricks. (The finger one I do see that way.) $\endgroup$ – Sue VanHattum Dec 16 '19 at 15:25
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I was taught the multiplication table for single digit numbers multipled by $9$ based on two observations: a. the digits of the result sum to $9$ b. the first digit of $k \times 9$ is $k-1$. This sort of trick is helpful because it makes the multiplication tables more interesting and easier to memorize. In particular the student feels proud for having mastered early the big number $9$.

For me as a child it was also one my first encounters with math (rather than arithmetic). I remember spending some time (probably with some parental urging) trying to understand why a. and b. are true, and why they are special to $9$.

Note that there are similar tricks for most of the multiplication tables - the digits of a multiple of $3$ sum to a multiple of $3$, the second digit of a multiple of $4$ is even. For integers less than $12$, only $7$ admits no tricks ...

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