## Integer Multiplication Explained

When I first left undergraduate school I took a job teaching 8th and 9th grade math. My 8th graders were introduced to integer multiplication for the first time at that level. They certainly knew natural number multiplication and even knew integer addition which includes subtraction by the time we took up integer multiplication.

When teaching integer addition I presented it as addition of vectors directed along the number line. So +2 + -4 would be an arrow starting at zero of length 2 pointing to the right then append an arrow of length 4 pointing to the left. The tip of the second arrow wound up at the point -2 on the number line.

To extend that idea to integer multiplication I introduced items I called ‘Number Line Frogs’. Each frog had a number associated with it and a direction. So a +2 frog could jump two places at a time and it pointed to the right. The number -3 was represented by a frog that could jump three places and pointed to the left.

I then asked the students to think of integer multiplication problems in two parts. The problem a x b was to be visualized as an ‘a-frog’ being asked to do something and that something was to perform ‘b’ jumps. The problem always started at the origin or ‘0’ on the number line. The sign of the number ‘b’ indicated which direction the frog was to jump. The absolute value of the number ‘b’ indicated how many jumps the frog was to take. The sign of the number ‘a’ indicated which direction the frog was pointing and the absolute value of the number ‘a’ indicated how far the frog could jump.

In this way of thinking about it the problem +2 x -3 was the same as saying ‘Allow a +2 frog to jump three jumps backward (-3)’. I stressed that the problem -3 x +2 was the same problem as could be verified by saying ‘Allow a -3 frog to take two jumps forward (+2)’ In both cases the end result was the number -6.

The rule for ‘positive-a times positive-b’ was to say ‘let a postive-a-frog take b-jumps-forward’ which came out to be the same problem as saying ‘let a positive-b-frog take a-jumps-forward’.

The rule for ‘positive-a times negative-b’ was to say ‘let a positive-a-frog take b-jumps-backward, while the rule for ‘negative-b times positive-a’ was to say ‘let a negative-b frog take a-jumps-forward’.

Finally the rule for ‘negative-a times negative-b’ was to say ‘let a ‘negative-a frog take ‘b-jumps-backward’.

When viewed this way and taking into account the ‘frog’ always started at zero the sign table for integer multiplication was justified. Additionally the commutative property of multiplication was illustrated by noting any number that could be decomposed into ‘b-groups of length a’ was equivalent to the observation that the same number could be decomposed into ‘a-groups of length b’.

I will say that the 8th grade group I had were sorted in an inhomogeneous manner with all the top students put into one class and that class was assigned to me. I did worry about this sorting process and felt that it was wrong on multiple levels but given there was nothing a first year math teacher could do to change the situation I made the most of it and presented many mathematical concepts using non standard methods (i.e. presented not-by-the-book). I think that math class did enjoy my presentations.

## Can I Get Your Digits! (Except for one of them)

I follow the blog “Math Jokes 4 Mathy Folks“.  A recent post gave a game/riddle you can play with kids or even post a set of questions on some social media site and play with your friends.  The post is “Can I Get Your Digits“.

So, here are the questions you can ask:

[Step-1] Pick a number with more than four digits (keep it secret).
[Step-2] Add the digits of the number together (keep it secret).
[Step-3] Subtract this sum from the original number (keep it secret).
[Step-4] From [Step-3] give me all the digits except for one of them (keep that digit secret).
[Step-5] I will ‘magically’ guess the missing (secret) digit!

The ‘magic-guess’ is: Add up the digits given to you and then add what ever number it takes to get to the nearest multiple of nine and that number is the missing digit.

So for example: the number 87694. When the digits are added you get 34.  When you subtract 36 from 87694 you get 87660.  Below is a table for each of the possible missing digits from number 87660.

Missing the ‘0’ … 8+7+6+6 = 27 (The missing digit is ZERO)
Missing the ‘6’ … 8+7+6+0 = 21 (The missing digit is SIX)
Missing the ‘7’ … 8+6+6+0 = 20 (The missing digit is SEVEN)
Missing the ‘8’ … 7+6+6+0 = 19 (The missing digit is EIGHT)

Now here is an explanation of why this always works.

The way numbers are represented is in what is called a ‘Place Value System’.  In this system a number can be thought of as a sum of the digits of the number times a power of 10 (for a base 10 number).  The power of 10 depends on the ‘Place’ of the digit.  The ‘least-significant’ digit (the on on the right end of a number) can be thought of as that digit times $10^{0}$.  The next ‘Place’ is called the 10’s place and can be thought of as that digit times $10^{1}$, with the next ‘Place’ being the 100’s place or the digit times $10^{2}$ and so forth.

So for example: $87694\quad =\quad 8\times {10}^{4}+7\times {10}^{3}+6\times {10}^{2}+9\times {10}^{1}+4\times {10}^{0}$

And this representation is equivalent to the following one when you replace the powers of 10 with the corresponding addition problem: $8\left( 1+9999 \right) +7\left( 1+999 \right) +6\left( 1+99 \right) +9\left( 1+9 \right) +4$

It is now clear how this ‘magic guess’ works. When you distribute each digit over the corresponding sum you get the following: $8+8\left( 9999 \right) +7+7\left( 999 \right) +6+6\left( 99 \right) +9+9\left( 9 \right) +4$

Regrouping you get the following very suggestive formula: $(8+7+6+9+4)\quad +\quad 8\left( 9999 \right) +7\left( 999 \right) +6\left( 99 \right) +9\left( 9 \right)$

This is clearly the sum of the digits of the original number plus another number that is clearly a multiple of nine: $\sum { digits\_ of(87694) } \quad +\quad 9\left[ 8\left( 1111 \right) +7\left( 111 \right) +6\left( 11 \right) +9\left( 1 \right) \right]$

The main thing to realize here is that when you subtract $\sum { digits\_ of(87694) }$ from the original number you are left with a number that (whatever its actual digits are) is a multiple of nine.

Finally it is the case that any number that is a multiple of nine also has the property that the number resulting from summing its digits is also a multiple of nine. This final fact is what you use to find the missing digit. Leaving out a single digit will ‘disturb’ the sum so it is no longer a multiple of nine and you only need sum up what you are given and find that number that causes the sum to be a multiple of nine to deduce the missing digit.

Notice: When the dropped digit is either a ‘0’ or a ‘9’ your magic ‘guess’ should be: “You left out either a 9 or a 0.”

Notice: This explanation can be extended to an arbitrary number of digits and thus it would be a proof.

Notice: That a multiple of nine also has the properties that the sum of its digits is also a multiple of nine can be explained in exactly the same way as the explanation above goes. So assume some number is a multiple of nine: $X\quad =\quad 9x$

Now write X as a sum of its digits with each one times its corresponding power of ten: $X\quad =\quad \sum _{ i=0 }^{ n }{ { a }_{ i } } { 10 }^{ i }$

Notice that we can write each term of the sum as a digit of the the original number times the quantity $\left( 1+9\cdots 9 \right)$ where the symbol “ $9\cdots 9$” is 9 or 99 or 999 or 9999 and so forth. $X\quad =\quad { a }_{ 0 }+\sum _{ i=1 }^{ n }{ { a }_{ i } } \left( 1+9\cdots 9 \right)$

This way of writing the number makes it clear what happens when each digit is distributed over the its corresponding sum and the the sum regrouped so that the sum of the digits of the original number are distinct from the rest of the number. $X\quad =\quad { a }_{ 0 }+\sum _{ i=1 }^{ n }{ { a }_{ i } } +\sum _{ i=1 }^{ n }{ { a }_{ i } } \left( 9\cdots 9 \right)$

Now it is clear that each term of the rightmost sum can be factored by the number nine giving the formula below. The symbol “ $1\cdots 1$” is 1 or 11 or 111 or 1111 and so forth. $X\quad =\quad { a }_{ 0 }+\sum _{ i=1 }^{ n }{ { a }_{ i } } +\quad 9\sum _{ i=1 }^{ n }{ { a }_{ i } } \left( 1\cdots 1 \right)$

Recalling that the original number is a multiple of nine we get the following: $9x\quad =\quad { a }_{ 0 }+\sum _{ i=1 }^{ n }{ { a }_{ i } } +\quad 9\sum _{ i=1 }^{ n }{ { a }_{ i } } \left( 1\cdots 1 \right)$

Finally subtracting the rightmost sum from both sides of the formula and factoring out the number nine we get the following which does show that the sum of the digits of the original number is a multiple of nine. $9\left[ x-\sum _{ i=1 }^{ n }{ { a }_{ i } } \left( 1\cdots 1 \right) \right] ={ a }_{ 0 }+\sum _{ i=1 }^{ n }{ { a }_{ i } } \quad$