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.

$9\left[ x-\sum _{ i=1 }^{ n }{ { a }_{ i } } \left( 1\cdots 1 \right) \right] ={ a }_{ 0 }+\sum _{ i=1 }^{ n }{ { a }_{ i } } \quad$

Fizz Buzz (Beep?)

I subscribe to a blog called Quora.  You can set filters on what sort of questions are presented. I’ve set them to show math, physics, and computer science posts.

Today I saw a post that asked if someone with an advanced degree in computer science could fail the “Fizz Buzz Test“.

I had no idea what this test was so I did a Google Search on the words “fizzbuzz test” and came up with a wikipedia page dedicated to the game which also mentions the test (here is the page).

After reading the article section named “Other Uses” and messing around with a python script today I can safely say I do fall into the category of someone who has used this game as a “Software Kata

Here is what I’ve done so far.  And, in my defense, I did start out with a VERY SIMPLE version of this game and it sort of “grew-like-topsy” to what you see below.

I’ve used the WordPress tags to embed the code below.  So this blog post not only talks about the Fizz Buzz Test but also gives a demonstration of how source code embedding works on WordPress.

# Fizz Buzz (beep) Script
import os

def sayIt( anIndex, aCount, phrase):
indexString = "The "+str(aCount)
if aCount >= 11 and aCount <= 20:
indexString += 'th'
elif aCount % 10 == 1:
indexString += 'st'
elif aCount % 10 == 2:
indexString += 'nd'
elif aCount % 10 == 3:
indexString += 'rd'
else:
indexString += 'th'
indexString += ' element is '+str(anIndex)+' and we say '+phrase
os.system("say " +indexString)

def findPhrase( index, fizzNumber, buzzNumber, beepNumber ):
newNumber = index
thePhrase = ''
if fizzNumber != 0:
while newNumber % fizzNumber == 0:
thePhrase += 'FIZZ '
newNumber = newNumber / fizzNumber
if buzzNumber != 0:
while newNumber % buzzNumber == 0:
thePhrase += 'BUZZ '
newNumber = newNumber / buzzNumber
if beepNumber != 0:
while newNumber % beepNumber == 0:
thePhrase += 'BEEP '
newNumber = newNumber / beepNumber
return thePhrase

def fizzBuzzBeep(maxCount, fizzNumber, buzzNumber, beepNumber):
count = 0
index = 1
while (count<maxCount):
aPhrase = findPhrase( index, fizzNumber, buzzNumber, beepNumber )
if len(aPhrase) != 0:
count +=1
print("{0:3d} - {1:3d} {2}".format(count, index, aPhrase))
sayIt( index, count, aPhrase)
index +=1

if __name__ == '__main__':
fizzBuzzBeep(100, 3, 5, 7)



The Volume of A Pizza

We ordered ‘Mellow Mushroom‘ pizzas for supper last night and we still have about one pie left over in the frig, just waiting to be finished off.

Now: What is pizza? one way to say it is to assume the thickness of the pie is the value $'a'$ and the radius of the pie is the value $'z'$ then to say, “What is Pizza?” could be to ask, “What is the volume of a pizza?”

Thinking of a pizza as a very thin cylinder of height $'h'$ and radius $'r'$ we get.

[1] $V = \pi r^{2}h$

Substituting $'a'$ for the height $'h'$ and $'z'$ for the radius $'r'$ we get:

[2] $V = \pi z^{2}a$

Next  recall that the greek letter $\pi$ can be written in English as $Pi$ we get:

[3] $V = Pi z^{2}a$

Finally expanding the $z{2}$ to be $z z$ we get:

[4] $V = Pi z z a$

Another Lottery Ticket

As you can see in the graphic below the numbers that were picked by the computer when I purchased the ticket did not match the numbers that were selected that evening.

So a question that occurred to me was: “How close did I get?” And one way to think about distances between two numbers is to subtract their values and take the absolute value of the result. So the distance between the number $2$ and the number $7$ would be $5$. Here is the formula:

$\lvert 2-7 \rvert = \sqrt{(2-7)^2} = \sqrt{(-5)^2} = \sqrt{25} = 5$

This is very easy to extend into higher dimensions. For example, if you go to two dimensions this becomes the pythagorean theorem as applied to two points in the plane. Here is the formula using the example of the points $(a,b) and (c,d)$

$dist( (a,b), (c,d) ) = \sqrt{(a-c)^2 + (b-d)^2}$

In the case of the lottery tickets we can think of a lottery ticket as a point in a six dimensional space, since the ticket has six numbers. This means that the distance between the ticket I purchased and the winning ticket is:

${\sqrt{(10-6)^2 + (13-8)^2 + (14-31)^2 + (22-46)^2 + (52-52)^2 + (11-29)^2 }}$

This expression simplifies to be $\sqrt{1230}$ which is about $35.1$, this means that I was pretty close when you consider the maximum distance between two sets of numbers in the lottery is ($125$). The maximum distance is computed by considering the distance between the two points $(1,2,3,4,5,1)$ and $(55,56,57,58,59,35)$ This type of distance is called the Euclidean Distance between two points..

E to the ‘i’ Pie

I’ve seen the following before but it is so cool I have to write it down:

The derivation of the formula below is really cool but this post is not about the derivation but an application of the formula:

[1]  $e^{ix} =$ $\cos x + i\sin x$

If we let $x = \pi$ we get

[2] $e^{i\pi} =$ $\cos \pi + i\sin \pi$

Simplify this by recalling that $\cos \pi = -1$ and $\sin \pi = 0$ we get:

[3] $e^{i\pi} = -1$

This is all really straight forward.  Here is the cool part.  What if we let $x = \pi/2$ then we get:

[4]  $e^{i\pi/2} =$ $\cos \pi/2 + i\sin \pi/2$

Since $\cos \pi/2 = 0$ and $\sin \pi/2 = i$ then this simplifies to be

[5]  $e^{i\pi/2} = i$

Finally if we raise both sides of the equation to the $i$ power we get:

[6] ${(e^{i\pi/2}})^i = i^i$

Combining the exponents on the left and noting that $(i)(i) =$ $i^2$ and $i^2 = -1$ we get

[7] $e^{-\pi/2} = i^i$

or

[8] $\dfrac{1}{e^{\pi/2}} = i^i$

or

[9] $\dfrac{1}{\sqrt{e^\pi}} = i^i$

Since the left hand side no longer has $i = \sqrt{-1}$ then if $e^{\pi}$ is a real number then $i^i$ is also a real number!   And, it is a real number see, for example, the articles “http://en.wikipedia.org/wiki/Gelfond%27s_constant” and “http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem

If you go to WolframAlpha.COM and enter the sting “evaluate $i^i$” it actually gives you a  number, marked as transcendental and it’s value is:

[10] $i^i =$ $0.207879576350761908546955 ...$

Finally I wrote this post after reading today “Saturday Morning Breakfast Cereal” cartoon.  WARNING: This day’s cartoon is most definitely N.S.F. (Not suitable for work!)

A Mathematicians Lottery

I follow the Spiked Math Blog and I especially liked the ‘Mathematicians Lottery‘ post which relies on the divergence of the harmonic series for the payout of the prize to insure that the payout can be managed.

The Wikipedia page on the harmonic series is very well done and the demonstation that this series diverges using an improper integral reminded me of the proof that I first saw when I was taking third semester calculs back in 1972 (forty years ago). Here is how the payout goes and the proof that you can award an arbitrary large sum of money as prizes in the lottery and get away with it even if you suppose that every ticket is a winner. Here is how all that goes:

[1] $\sum_{n=1}^{\infty}\frac{1}{n}$

The idea is that payout works like this: You get $1.00 the first week then (1/2)($1.00) the second week then (1/3)($1.00) the third week and because this series is divergent its partial sum is eventually greater than any finite number (even$750,000,000,000,000 “750 trillion dollars”).

The key is that the harmonic series diverges very, very slowly. To see that the series diverges at all it is enough to notice that it is bounded (from below) by something else that you can show ‘goes-to-infinity’. The choice is the following integral:

[2] ${\int^{\infty}_1\frac{1}{n}\,dn}$

To see why the integral bounds the series from below this picture is a great help:

Since the area under the curve from 1 to infinity is strictly less than the sum of the areas of the rectangles then if we can show that the integral diverges then that will show that the series diverges. This is because each of the rectangles has base equal to 1 and height equal to 1/n and thus area equal to 1/n and the sum of them all is exactly the value of the infinite series.

When you actually perform the integration in [2] you get:

[3] ${\int^x_1\frac{1}{n}\,dn = ln(x)-ln(1)}$

To see why this integrates to ln(x) you think about the inverse function theorem and the fact that the derivative of exponential is itself and the inverse of the exponential is the lograthim.

[4] ${\frac{d}{dx}\left(e^x\right) = e^x}$

and

[5] if ${f(x) = e^x}$ then ${f^{-1}(x) = ln(x)}$

This also explains why $ln(x)$ is an increasing function and why it is a very, very slowly increasing function (since its inverse is a very very quickly increasing function). And all this says that:

[6] $\sum_{n=1}^{\infty}\frac{1}{n} > {\lim_{x\to\infty}\int^x_1\frac{1}{n}\,dn = \lim_{x\to\infty}ln(x) = \infty}$

So this settles it, the harmonic series is divergent and you can always exceed an arbitrary large by just picking a greater and greater partial sum. The question then comes up just how many years do you need to wait to collect 750 trillion dollars. This comes out to be:
[7] $ln(n) = \frac{(7.5)(10^{14})}{52}$ because there are 52 weeks in a year.

Now we just solve for n getting:

[8] $n = e^{\frac{(7.5)(10^{14})}{52}}$

I did visit Wolfram Alpha web site and entered the exponential and, of course, it is way beyond computation, but interestingly Wolfram Alpha displayed the following:

[9] $e^{\frac{(7.5)(10^{14})}{52}} = 10^{10^{10^{1.119224798480347}}}$

And this totally settles the matter as to whether or not it is OK award 750 trillion dollars to each ticket holder.