Education Matters

10 Jul

Earth: Hotter than Hell – What!

Author: Administrator // Category: Maths

On March 25, 2010, Newsweek Religion Editor Lisa Miller appeared on the National Public Radio program, The Diane Rehm Show. Diane was speaking with Lisa about Lisa’s latest book Heaven: Our Enduring Fascination with the afterlife.

Listening to Lisa talk about people’s diverse views of Heaven I got to wondering a few things about Heaven. I wondered, Where was Heaven? How far was Heaven from my house? What do people do in Heaven? What do the souls in Heaven look like and what do they do there? What is the average temperature in the Heaven? The last question, the temperature question, involved a measurement and was, I thought, the only one to which I might be able to find an answer. (It turns out I could not, but I did come up with an answer to a related, still very interesting question.) Being familiar (and only vaguely so) with the Christian religion, I narrowed my question to be What is the average temperature in the Christian Heaven?

The Average Temperature of Heaven

In my searches through Bible literature I could not find a direct answer to this question, but along the way to the answer I decided, through some straight-forward computations, that the temperature on Earth could be higher than the temperature in Hell and people would still be healthy, happy, and prospering.

Earth – Hotter than Hell

If you believe in the Christian concepts of Heaven and Hell, then proof of this fantastic statement can be found in the Bible. Here is how it goes.

We’ll use three facts, two as stated in The Bible, and one from the well-known law from Earthly physics, the The Stefan–Boltzmann law.

Bible Facts

We’ll take the two Bible facts from two sources, the King James Version and the New International Version.

Here are the verses from both versions followed by an interpretation of those verses. (found at COLLECTIONS Commentaries, Word Studies, Devotionals, Sermons, Illustrations Old and New Testament).

Isaiah 30:23-26 can be interpreted to tell us how much light will be on Earth when Christ establishes His Millennial Reign upon the earth.

Isaiah 30:23-26, The King James Version

The Book of Isaiah

[23] Then shall he give the rain of thy seed, that thou shalt sow the ground withal; and bread of the increase of the earth, and it shall be fat and plenteous: in that day shall thy cattle feed in large pastures. [24] The oxen likewise and the young asses that ear the ground shall eat clean provender, which hath been winnowed with the shovel and with the fan. [25] And there shall be upon every high mountain, and upon every high hill, rivers and streams of waters in the day of the great slaughter, when the towers fall. [26] Moreover the light of the moon shall be as the light of the sun, and the light of the sun shall be sevenfold, as the light of seven days, in the day that the LORD bindeth up the breach of his people, and healeth the stroke of their wound.

Isaiah 30:23-26, the New International Version

[23] He will also send you rain for the seed you sow in the ground, and the food that comes from the land will be rich and plentiful. In that day your cattle will graze in broad meadows. [24] The oxen and donkeys that work the soil will eat fodder and mash, spread out with fork and shovel. [25] In the day of great slaughter, when the towers fall, streams of water will flow on every high mountain and every lofty hill. [26] The moon will shine like the sun, and the sunlight will be seven times brighter, like the light of seven full days, when the LORD binds up the bruises of his people and heals the wounds he inflicted.

An Interpretation We see that the curse on the earth and over Jerusalem is being removed by the Lord because: The rain is falling and causing good growth of the crops. The yield of the ground is rich and plenteous. The livestock are grazing in a roomy pasture. Every loft mountain and high hill has streams running with water. The wicked have been slaughtered and removed from the earth. The light of the moon is as the light of the sun, and the light of the sun is seven times brighter than normal.The fulfillment of these promises concerning the land of Israel has in preparation of Christ’s Millennial Reign already been occurring. When Israel came into the land in 1948 and was declared a nation again, the land was deserted and considered completely worthless and uninhabitable, and no one had wanted the land. Today, however the Lord has blessed the land and it produces much agriculture and fruits, and much of it has even been reforested. This is the first desert land upon the earth which has ever been refurbished, everywhere else in the world desert has always been constantly advancing.

It looks like things will be pretty nice on Earth. Except, as announced, but hidden, in Verse 26, it will be hotter on Earth than it is in Hell.

The Temperature on Earth as Given by Isaiah 30:26, the Mathematics

Our clue as to the temperature of Earth is in verse 26, The moon will shine like the sun, and the sunlight will be seven times brighter, like the light of seven full days…

Light is electromagnetic radiation. According to verse 26, the amount of radiation received by the Earth is (will be)

7 times 7 = 49 times as much radiation as that contributed by the moon.

So, the Earth will receive 1 (moon) + 49 (suns) = 50 times its current amount of radiation.

Now we’ll employ the Stefan-Boltzman Law. This law expresses, mathematically, the relationship between amount of radiation and temperature.

The Stefan-Boltzman Law states that the energy radiated by a radiator (the Earth in our case) per second per unit area is proportional to the fourth power of the body’s absolute temperature. Using

$E_{new}$ to represent the new temperature of the Earth,

$E_{orig}$ to represent the original temperature of the Earth,

we get

$\displaystyle{\Biggl(\frac{E_{new}}{E_{orig}}\Biggr)}^4 = 50$

We can solve this equation to obtain the new temperature of the Earth. We will solve for $E_{new}$ .

$\begin{array}{clclcl} \displaystyle{\Biggl(\frac{E_{new}}{E_{orig}}\Biggr)}^4 &= 50 \cdot t \\ \displaystyle{\frac{E_{new}^4}{E_{orig}^4}} &= 50 \\ E_{new}^4 &=50 \cdot E_{orig}^4 t \\ E &= \root 4 \of 50 \cdot E_{orig} \end{array}$

So, the new temperature of the Earth will be $\root 4 \of 50$ times the original temperature of the Earth. The temperature of the Earth has been established to be $279^\circ$ K.

Then,

$\begin{array}{clclcl} E_{new} &= \root 4 \of 50 \cdot E_{orig} \\ E_{new} &= \root 4 \of 50 \cdot 279 \\ E_{new} &= 742 \end{array}$

The new temperature of Earth is, or will be, $742^\circ$ K. This Kelvin temperature translates to $469^\circ$ C and $876^\circ$ F. Geez! $876^\circ$ F is really hot.

The Temperature of Hell

We cannot get the temperature of Hell as precisely as we did the newer temperature of the Earth. But we can get an interval estimate for it. The interval comes from information contained in The Book of Revelations.

Revelations 21:8 tells us
But the fearful, and unbelieving, and the abominable, and murderers, and whoremongers, and sorcerers, and idolaters, and all liars, shall have their part in the lake which burneth with fire and brimstone: which is the second death.

Revelation 21:8 (New International Version)

The Book of Revelation

But the cowardly, the unbelieving, the vile, the murderers, the sexually immoral, those who practice magic arts, the idolaters and all liars—their place will be in the fiery lake of burning sulfur. This is the second death.”

Brimstone is another name for sulfur. According to the New World Encyclopedia, sulfur melts at 239.38 °F (388.36 K, 115.21 °C). This means that the temperature of Hell must be at least 239.38 °F. Also, the boiling point of sulfur is 832.3 °F (717.8 K, 444.6 °C). At temperatures higher than 832.3 °F, sulfur boils and turns to a gas and is no longer sulfur. This means that the maximum temperature of Hell is 832.3 °F .

Now we have that the newer temperature of Earth is $876^\circ$ F and the maximum temperature of Hell is 832.3 °F.

So there it is, expressed in the Bible — Earth will be hotter than Hell.

Since people have lived on the Earth, it has never been $876^\circ$ F. Because the current temperature of Earth is not 876 °F, we can conclude that the curse on the earth and over Jerusalem has not been removed by the Lord. Otherwise, we have a contradiction. So it must be sometime in the future that the curse on the earth and over Jerusalem will be removed by the Lord.

But even at $876^\circ$ F, Isaiah 30:25-25 states that all people will be healthy and prospering.

So now I wonder if the bodies of people suddenly change to tolerate this new, high temperature, or if they will evolve to tolerate it.

I wonder, too, about the temperature of Heaven. I think Heaven must be cooler than Hell, else people would burn there. So maybe an upper bound on the temperature of Heaven is 832.3 °F, the boiling point of brimstone.

Or, maybe Isaiah got the whole number thing wrong.

Or, maybe I got the numbers wrong. It can happen. One time in a calculus II class, I presented a very nice derivation of the the coordinates of the center of mass of a thin sheet, but then realized I got them exactly backwards. $(\bar{x}, \bar{y}) \leftrightarrow (\bar{y}, \bar{x})$ . Geez. I am glad I am not in charge of temperatures.

I found quite a few articles on the internet relating the temperatures of Heaven and Hell. All that I found conclude that Heaven is Hotter than Hell. They do so by interpreting Isaiah 30:26 as giving the amount of radiation contained in Heaven. I think that the previous lines, 23-25, tell us that the amount of radiation is the amount received on the Earth, not the amount contained in Heaven. That’s my take, anyway.

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06 Jul

Copernicium and Nutrium

Author: Administrator // Category: Maths

I read with great excitement on Yahoo News about the official naming of the heaviest element known to humankind. The Yahoo article, dated February 24, 2010, proclaimed that Copernicium was named after the Polish astronomer Nicolaus Copernicus and is 277 times as heavy as the lightest known element, hydrogen.

I called my Polish friends from my college’s mathematics department and asked them to meet for breakfast at MiMi’s, a nice local Polish Cafe near the school, to discuss this wonderful event.

I made a short video to capture our joy. In the video,

1. Jim Vilchuck appears on the left. Jim is real smart guy who likes to talk in detail about both the coefficient of determination and the coefficient of variation. Start any discussion with him and within 5 minutes he’ll have you asking about $r^2$ . You can’t help yourself. Soon you will find yourself asking Jim about other coefficients. He has a remarkable talent.

2. Sven Svenkoski appears in the center of the picture. Sven, at the age of 10, while practicing some topological probability homework problems, discovered a submarine lost for 30 years some 80,000 ft down on the Atlantic Ocean floor. Sven’s discovery came just in time as the crew of 150 sailors was just about out of oxygen. When we started discussing Copernicium, Sven suggested the heaviest element was being named after Copernicus because it was Copernicus who wrote the 1969 hit song, He Ain’t Heavy, He’s my Brother. Sven has an Erdos Number of 1!

3. Craig Allenski appears on the right in the picture. Craig is one of the smartest people in the department. I remember going to him when I had questions about how to derive $\bar{x}$ and $\bar{y}$ when presenting the topic Center of Mass in calculus. For some reason I still don’t understand, one morning Craig beat me up in the school parking lot.

See the guys discussing Copernicium in this video clip. The guys discussing Copernicium.

We had a nice breakfast, spent about 2 hours talking about mathematics, then went our own ways.

It was raining pretty hard the morning of our breakfast and as I hurried to my car from the restaurant, I got pretty wet. I was still pretty wet when I got home some 10 minutes later. In the bathroom while blow drying my hair, I saw something that just astonished me. See Figure 1.

Figure 1 A bottle of Nutrium

Do you see this? Nutrium. An element I had never heard of. Not only had I not heard of it, but the Dove company was using it in soap! Now I am pretty good at keeping up with the Periodic Table, but somehow I had completely missed the discovery of Nutrium. But there it was with its electron configuration right there on the bottle of Dove soap. See the Periodic Table picture.

The Periodic Table

I had to know more about this element, Nutrium. So just like anybody would, I put a squirt of the soap on a glass slide and took it to my microscope lab.

My house is pretty much like most people’s house, I have the standard living room, kitchen, bedrooms, bathrooms, and hallways leading to other rooms likes laboratories and engine rooms. See Figure 2.

Figure 2 Main hallway in Denny’s house

I put the slide containing the Nutrium under the microscope to get a visual of this element. See Figure 3.

Figure 3 Soap sample

An Astonishing Discovery

Among all the other particles in the sample, I was able to detect the Nutrium atom. Figure 4 shows an individual Nutrium atom.

Figure 4 The Nutrium atom

I zoomed into one of outer electrons. Figure 5 shows the zoom magnification at 10 kabillion times the atom’s original size. Look at the astounding electron at the top of the image! Do you see it? That electron looks like a guy in a 60’s surf music rock band! What the heck, I shouted!

Figure 5 Nutrium atom with the Ellis electron characteristic

I connected the microscope to speakers so as to get an audio of the Nutrium atom.

Speakers connecting to the microscope

The Nutrium Atom – See it and Hear it Here

What a surprise! You have got to see and hear this element. Be ready to adjust the volume on your computer’s speakers.

Here it is, The sight and sound of Nutrium

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25 May

Don T’s Range of Motion

Author: Administrator // Category: Maths

A guy I know, Don Thomas (known to his family as Donut Head), told me that just recently he hurt his wrist and that its range of movement was only about 20% its normal range of movement. He went on to say that he thought his wrist function was improving at a rate of 1% a day.

So how long, Don T asked me, will it take until my range is back to 100%?

Assuming that he, that would be Don T, was down to 20% his full range of motion, we calculated as follows.

We let $F$ represent Don’s full range of motion.

Then, 20% of full range can be represented by $0.20F$ . Its true. Think about it this way and just write what you say.

20% of current range.
The word of translates into mathematics as times $\cdot$ .
Since the current range is $F$ , 20% of $F$ is represented as $0.20\cdot F$ , or just $0.20F$ .

You know, I think I will use the star symbol $\star$ rather than the multiplication dot $\cdot$ to indicate multiplication. With decimal points and multiplication dots, there are too many dots.

We called the day of injury Day 0.
The amount of motion on Day 0 is $0.20F$ .

Now, we will let $A_n$ represent the total amount of motion at the end of Day $n$ .

Then, on Day 0, the total amount of motion is

$A_0 = 0.20F$

We called the first day following Day 0, Day 1.
A 1% increase in range is represented by $0.01\star \{\hbox{the current amount of range}\}$ .

Since the current range is $0.20F$ , a 1% increase is represented as $0.01 \star 0.20F$ .

Then, the amount of motion at the end of Day 1 is the range of motion on Day 0 plus the 1% increase in that range. That is, the total amount of motion at the end of Day 1 is

$A_1 = 0.20F + 0.01 \star 0.20F$ .

We can simplify a little by factoring out the $0.20F$ that is common to the two terms on the right side of the formula. Doing so gives us

$\begin{aligned} A_1 &=& 0.20F + 0.01 \star 0.20F \hfill \\ A_1 &=& 0.20F \bigl(1 + 0.01\bigr) \hfill \\ A_1 &=& 0.20F \bigl(1.01\bigr)\hfill \end{aligned}$

We can use exponents to related the day number and the total amount of motion at the end of that day.

At the end of Day 0, the total amount of motion $A_0 = \bigl(0.20F\bigr)^0$ .

At the end of Day 1, the total amount of motion $A_1 = 0.20F \bigl(1.01\bigr)^1$ .

We called the first day following Day 1, Day 2.
A 1% increase in range is represented by $0.01\star \{\hbox{the current amount of range at the end of Day 1}\}$ .

Since the amount of range at the end of Day 1 is $0.20F \bigl(1.01\bigr)^1$ , a 1% increase is represented as $0.01 \star 0.20F \bigl(1.01\bigr)$ .

Then, the amount of motion at the end of Day 2 is the range of motion at the end of Day 1 plus the 1% increase in that range. That is, the total amount of motion at the end of Day 2 is

$A_2 = 0.20F \bigl(1.01\bigr)^1 + 0.01 \star 0.20F \bigl(1.01\bigr)$ .

We can simplify a little by factoring out the $0.20F \bigl(1.01\bigr)$ that is common to the two terms on the right side of the formula. Doing so gives us

$\begin{aligned} A_2 &=& 0.20F \bigl(1.01\bigr)^1 + 0.01 \star 0.20F \bigl(1.01\bigr) \\ A_2 &=& 0.20F \bigl(1.01\bigr) \bigl(1 + 0.01\bigr) \\ A_2 &=& 0.20F \bigl(1.01\bigr)^2 \end{aligned}$

You may be able to see the general form, but if not, we’ll make one more iteration.

At the end of Day 0, the total amount of motion $A_0 = \bigl(0.20F\bigr)^0$ .

At the end of Day 1, the total amount of motion $A_1 = 0.20F \bigl(1.01\bigr)^1$ .

At the end of Day 2, the total amount of motion $A_2 = 0.20F \bigl(1.01\bigr)^2$ .

We called the first day following Day 2, Day 3.
A 1% increase in range is represented by $0.01\star \{\hbox{the current amount of range at the end of Day 2}\}$ .

Since the amount of range at the end of Day 2 is $0.20F \bigl(1.01\bigr)^2$ , a 1% increase is represented as $0.01 \star 0.20F \bigl(1.01\bigr)^2$ .

Then, the amount of motion at the end of Day 3 is the range of motion at the end of Day 2 plus the 1% increase in that range. That is, the total amount of motion at the end of Day 3 is

$A_1 = 0.20F \bigl(1.01\bigr)^2 + 0.01 \star 0.20F \bigl(1.01\bigr)$ .

We can simplify a little by factoring out the $0.20F \bigl(1.01\bigr)$ that is common to the two terms on the right side of the formula. Doing so gives us

$\begin{aligned} A_3 &=& 0.20F \bigl(1.01\bigr)^2 + 0.01 \star 0.20F \bigl(1.01\bigr) \\ A_3 &=& 0.20F \bigl(1.01\bigr)^2 \bigl(1 + 0.01\bigr) \\ A_3 &=& 0.20F \bigl(1.01\bigr)^3 \end{aligned}$

Now we have it. We can see the general form.

The total amount of motion at the end of Day $n$ is given by
$A_n = 0.20F \bigl(1.01\bigr)^n$

We can use this formula to determine how many days it will take to reach 100% range.

100% of the total range is represented by $1.00F$ , or $1F$ .

We want to find the number of days, $n$ , it takes to get to $1F$ .

Set $A_n$ equal to $1F$ in the formula $A_n = 0.20F \bigl(1.01\bigr)^n$ and solve for $n$ .

$1F = 0.20F \bigl(1.01\bigr)^n$

Divide both sides by $F$ .

$\displaystyle{\frac{1F}{F}} = \displaystyle{\frac{0.20F \bigl(1.01\bigr)^n}{F}}$

The $F$ ’s divide out (cancel), and we are left with

$1 = 0.20 \bigl(1.01\bigr)^n$

Divide both sides by 0.20.

$\frac{1}{0.20} = \frac{0.20 \bigl(1.01\bigr)^n}{0.20}$

Which gives us

$5 = \bigl(1.01\bigr)^n$

To solve for $n$ we will take the natural log of each side.

$\begin{aligned} \ln(5) &=& \ln\bigl(1.01\bigr)^n \\ \ln(5) &=& n\cdot \ln(1.01) \\ n &=& \frac{\ln(5)}{\ln(1.01)} \end{aligned}$

In decimal form, $n=~162$ days.

Now, Donut Head’s doctor told him his wrist would take about 60 days to get back to 100%. So, that constant 1% increase in range of motion must be incorrect or he started with more than 20% his original range. Let’s assume he really did start with 20% his original range.

Then the problem is, given 100% range of motion in 60 days, that is, given $n = 60$ , find the rate at which the range of motion increases. Can you do it?

Can you solve $5 = \bigl(1 + r\bigr)^{60}$ for $r$ ?

You don’t need logs.

I got $r \approx 2.7\%$

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09 Feb

I’ll Follow the Graph

Author: Administrator // Category: Maths

Way back in time on March 8, 2009, I posted the blog article Note Frequencies in Jimmy Buffet’s Come Monday. In that article I computed and determined the average note frequency in Jimmy’s song Come Monday. By average note frequency I mean the average frequency in string vibrations measured in Hertz (Hz). One string vibration per second equals 1 Hz. Higher pitched notes have high Hertz values and lower pitched notes have low Hertz values. I noted in that article that the average (mean) note frequency of Come Monday was 319.78 Hz and used a frequency generator to produce the sound associated with that frequency. If you are reading this article you probably don’t have much to do, so check out the Come Monday article at http://www.dennymath.com/?p=351.

Since I posted that article, I have been thinking about connections between mathematics and music. I started wondering if songs might have equations associated with them and if so, what the graphs of those equations would look like. The thought didn’t seem that strange to me since sheet music looks somewhat graphical. As you look at a song’s sheet music, dots rise and fall, just like one sees when looking at the graph of discrete data points.

Because my niece Randi Jean loves Beatles music and because she can exert great influence on cosmic events, I decided to try to graph a Beatles song. I bought a book of Beatles sheet music and decided to see if I could create a graph of the song I’ll Follow the Sun.

Randi Jean listening intensely to a Beatles tune

You can see a youtube version of I’ll Follow the Sun as performed by Paul McCartney at

http://www.youtube.com/watch?v=LfO1nbCX0g8&feature=related

The Process

1. I copied the music for the song on my printer.

2. The song is in 4/4 time and the shortest note in the song is an 1/8 note. So, I decided to count 8 beats to each measure.

3. From wikipedia I copied a nice picture of notes so that I would have an idea of each note’s frequency value. See Figure 1.

http://en.wikipedia.org/wiki/Musical_notes

Figure 1 – Note frequency vs note name

4. Using Excel, I made a list of the notes and their corresponding frequencies. See Figure 2. Click on the picture to see it larger.

Figure 2 – Notes and their frequencies

5. Next to each note, I wrote in red ink the note’s frequency in Hertz (Hz).

6. Then next to each note I wrote in blue ink the number of beats the note required to be played.

Figures 3 and 4 show the a cropped version of the sheet music and my notation. Click on the picture to see it larger.

Figure 3 – Verse 1

Figure 4 – Refrain

The following steps show how I created the lists of data, the scatter diagram, the coordinate system, the piecewise function, and the graph of the data using the T-Nspire. They are just a how-to-do-this type of thing. If you want to skip it, scroll down to the The Graph.

7. I accessed my copy of the new TI-Nspire software that emulates the Texas Instrument’s TI–Nspire calculator.

8. Using Nspire, I first created a list of values. The first column in my list is the beat number, and the second column, the note frequency. See Video 1.

Video 1

9. Then, still using Nspire, I created a scatter diagram of my data. See Video 2.

video 2

10. Still using the Nspire, I created a piecewise linear function of the data in the beat vs Hz lists. See Video03.

Video 3

11. Video04 shows a summary of steps 8-10.

Video 4

12. Video 5 shows a minor change in the horizontal axis, one that makes the graph look a little better.

Video 5

13. The following four figures shows the four piecewise functions I constructed to create the graph.

Function, $f$ 1 produces verse 1.

Function $f$ 2 produces verse 2.

Function $f$ 3 produces the first part of the refrain, and

Function $f$ 4 produces the last part of the refrain.

The Graph

Figure 5 shows a black and white version of the graph of I’ll Follow the Sun when plotted as a piecewise function.

Figure 5

Figure 6 shows a colorized view of the graph to make it easier to see where the verses and refrain start and end. Click on the picture to make it larger.

Figure 6

I don’t know yet how to label axes on the Nspire. I will work on that.

I think what I will try next is to graph another Beatles song, or even a few more Beatles songs of approximately the same size, (number of measures) and compare them to each other. I am not sure just yet how I will compare them, but my colleague Joe suggested I use a subtraction method to see how various differences compare with each other.

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08 Feb

Sari Dates

Author: Administrator // Category: Maths

My daughter Sandy’s friend, Sari, noted on December 16, 2009, that on that particular day, when the date was expressed as 12/16/09, the square of the month equaled the product of the day of the month and the year. That is,

December squared = (the 16th) times (09), or better yet,

$12^2 = 16 \cdot 9$

I think this is a fun relationship and will give such dates the name Sari Dates. So, of course, we need to know how many Sari Dates there are.

I counted 40 of them. I listed each month and just counted. Here is how.

In general we want to count the number of occurrences of the event

$\hbox{month}^2 = \hbox{day} \cdot \hbox{year}$

Let’s narrow it down to this century.

I am just going to make a straight count.

Let $a$ = the month number
$b$ = day number
$c$ = year number

So, we are looking for how many times between 2000 and 2099 that

$a^2 = b\cdot c$

Not all numbers work. There are some restrictions on the values $a$ , $b$ , and $c$ .

Because $a$ represents a month, it must be between and including 1 and 12. That is,

Because $b$ represents a day, it must be between and including 1 and 28, or 1 and 29, or 1 and 30, or 1 and 31.

Because $c$ represents a year in this century, it must be between and including 0 (2000) and 99 (2099).

To make the count, start with the square of the month then find all the factors of that number that meet the

Let’s count.

January:
$a^2 = b\cdot c$

$01^2 = b\cdot c$

$1 = b\cdot c$

The only factors of 1 are 1 and 1. So,

$b=1, c=1$

Then for January, we get only one date, 01/01/01

February:
$a^2 = b\cdot c$

$02^2 = b\cdot c$

$4 = b\cdot c$

The factors of 4 are 4,1; 2,2. So

$b=4, c=1 \rightarrow 2^2 = 4 \cdot 1$

$b=1, c=4 \rightarrow 2^2 = 1 \cdot 4$

$b = 2, c=2 \rightarrow 2^2 = 2 \cdot 2$

Then for February, we get 3 dates: 02/02/01, 02/02/02, and 01/01/04.

March:
$a^2 = b\cdot c$

$03^2 = b\cdot c$

$9 = b\cdot c$

The factors of 9 are 9,1; 3,3. So

$b=9, c=1 \rightarrow 3^2 = 9 \cdot 1$

$b=1, c=9 \rightarrow 3^2 = 1 \cdot 9$

$b = 3, c=3 \rightarrow 3^2 = 3 \cdot 3$

Then for March, we get 3 dates: 03/09/01, 03/03/03, and 01/01/09.

April:
$a^2 = b\cdot c$

$04^2 = b\cdot c$

$16 = b\cdot c$

The factors of 16 are 16,1; 8,2; 4,4. So

$b=16, c=1 \rightarrow 4^2 = 16 \cdot 1$

$b=1, c=16 \rightarrow 4^2 = 1 \cdot 16$

$b=8, c=2 \rightarrow 4^2 = 8 \cdot 2$

$b=2, c=8 \rightarrow 4^2 = 2 \cdot 8$

$b=4, c=4 \rightarrow 4^2 = 4 \cdot 4$

Then for April, we get 5 dates: 05/16/01, 04/08/02, 04/04/04, 04/02/08, and 04/01/16.

May:
$a^2 = b\cdot c$

$05^2 = b\cdot c$

$25 = b\cdot c$

The factors of 25 are 25,1; 5,5. So

$b=25, c=1$ But this cannot happen because we can’t have a day of 25!

$b=1, c=25 \rightarrow 5^2 = 1 \cdot 25$

$b=5, c=5 \rightarrow 5^2 = 5 \cdot 5$

Then for May, we get 2 dates: 05/05/05, and 05/01/25.

June:
$a^2 = b\cdot c$

$06^2 = b\cdot c$

$36 = b\cdot c$

The factors of 36 are 36,1; 18,2; 9,4; 3,12. So

$b=36, c=1$ But this cannot happen because we can’t have a day of 36!

$b=1, c=36 \rightarrow 6^2 = 1 \cdot 36$

$b=18, c=2 \rightarrow 6^2 = 18 \cdot 2$

$b=2, c=18 \rightarrow 6^2 = 2 \cdot 18$

$b=9, c=4 \rightarrow 6^2 = 9 \cdot 4$

$b=4, c=9 \rightarrow 6^2 = 4 \cdot 9$

$b=3, c=12 \rightarrow 6^2 = 3 \cdot 12$

$b=12, c=3 \rightarrow 6^2 = 12 \cdot 2$

Then for June, we get 6 dates: 06/18/02, 06/09/04, 06/04/09, 06/03/12, 06/02/18, and 06/01/36.

July:
$a^2 = b\cdot c$

$07^2 = b\cdot c$

$49 = b\cdot c$

The factors of 49 are 49,1; 7,7. So

$b=49, c=1$ But this cannot happen because we can’t have a day of 49!

$b=1, c=49 \rightarrow 7^2 = 1 \cdot 49$ B

$b=7, c=7 \rightarrow 7^2 = 7 \cdot 7$

Then for July, we get 6 dates: 06/18/02, 06/09/04, 06/04/09, 06/03/12, 06/02/18, and 06/01/36.

August:
$a^2 = b\cdot c$

$08^2 = b\cdot c$

$64 = b\cdot c$

The factors of 64 are 64,1; 32,2; 16,4, 8,8. So

$b=64, c=1$ But this cannot happen since we can’t have a day of 64!

$b=1, c=64 \rightarrow 8^2 = 1 \cdot 64$

$b=32, c=2$ But this cannot happen since we can’t have a day of 32!

$b=2, c=32 \rightarrow 8^2 = 2 \cdot 32$

$b=16, c=4 \rightarrow 8^2 = 16 \cdot 4$

$b=4, c=16 \rightarrow 8^2 = 4 \cdot 16$

$b=8, c=8 \rightarrow 8^2 = 8 \cdot 8$

Then for August, we get 5 dates: 08/16/04, 08/08/08, 08/04/16, 08/02/32, 08/01/64.

September:
$a^2 = b\cdot c$

$09^2 = b\cdot c$

$81 = b\cdot c$

The factors of 81 are 81,1; 9,9. So

$b=81, c=1$ But this cannot happen since we can’t have a day of 81!

$b=1, c=81 \rightarrow 9^2 = 1 \cdot 81$

$b=9, c=9 \rightarrow 9^2 = 9 \cdot 9$

Then for September, we get 2 dates: 09/09/09 and 09/01/64.

October:
$a^2 = b\cdot c$

$10^2 = b\cdot c$

$100 = b\cdot c$

The factors of 100 are 100,1; 50,2; 25,4; 20,5; 10,10. So

$b=100, c=1$ But this cannot happen since we can’t have a day of 100!

$b=1, c=100$ But this cannot happen since we can’t have a year of 100! That you would be the year 3000.

$b=50, c=2$ But this cannot happen since we can’t have a day of 50!

$b=2, c=50 \rightarrow 10^2 = 2 \cdot 50$

$b=25, c=4 \rightarrow 10^2 = 25 \cdot 4$

$b=4, c=25 \rightarrow 10^2 = 4 \cdot 25$

$b=20, c=5 \rightarrow 10^2 = 20 \cdot 5$

$b=5, c=20 \rightarrow 10^2 = 5 \cdot 20$

$b=10, c=10 \rightarrow 10^2 = 10 \cdot 10$

Then for October, we get 6 dates: 10/25/04, 10/20/05, 10/10/10, 10/05/20, 10/04/25, and 10/02/50.

November:
$a^2 = b\cdot c$

$11^2 = b\cdot c$

$121 = b\cdot c$

The factors of 121 are 121,1; 11,11. So

$b=121, c=1$ But this cannot happen since we can’t have a day of 121!

$b=1, c=121$ But this cannot happen since we can’t have a year of 121!

$b=11, c=11 \rightarrow 11^2 = 11 \cdot 11$

Then for November, we get 1 date: 111/11/11.

December:
$a^2 = b\cdot c$

$12^2 = b\cdot c$

$144 = b\cdot c$

The factors of 144 are 144,1; 72,2; 48,3; 36,4; 24,6; 18,8; 16,9; and 12,12. So

$b=144 c=1$ But this cannot happen since we can’t have a day of 144!

$b=1, c=144$ But this cannot happen since we can’t have a year of 144!

$b=72, c=2$ But this cannot happen since we can’t have a day of 72!

$b=2, c=72\rightarrow 12^2 = 2 \cdot 72$

$b=48, c=3$ But this cannot happen since we can’t have a day of 48!

$b=3, c=48 \rightarrow 12^2 = 2 \cdot 48$

$b=36 c=4$ But this cannot happen since we can’t have a day of 36!

$b=4, c=36 \rightarrow 12^2 = 4 \cdot 36$

$b=24, c=6 \rightarrow 12^2 = 24 \cdot 6$

$b=6, c=24 \rightarrow 12^2 = 6 \cdot 24$

$b=18, c=8 \rightarrow 12^2 = 18 \cdot 8$

$b=8 c=18 \rightarrow 12^2 = 8 \cdot18$

$b=16, c=9 \rightarrow 12^2 = 16 \cdot 9$

$b=9, c=16 \rightarrow 12^2 = 9 \cdot 16$

$b=12, c=12 \rightarrow 12^2 = 12 \cdot 12$

Then for December, we get 7 dates: 12/24/06, 12/16/09, 12/18/08, 12/12/12, 12/09/16, 12/06/24, and 12/03/48.

To summarize, letting SD stand for Sari Date, I counted:

January: 1 SD
February: 3 SDs
March: 3 SDs
April: 5 SDs
May: 2 SDs
June: 6 SDs
July: 2 SDs
August: 5 SDs
September: 2 SDs
October: 3 SDs
November: 1 SD
December: 7 SDs

Then,

$1\hbox{SD} + 3\hbox{SD} + 3\hbox{SD} + 5\hbox{SD} + 2\hbox{SD} + 6\hbox{SD} + 2\hbox{SD} + 5\hbox{SD} + 2\hbox{SD} + 3\hbox{SD} + 1\hbox{SD} + 7\hbox{SD} = 40\hbox{SD}$

Thank you very much. This century, I count 40 Sari Dates.

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