### Number 243 of a series

Each letter represents a digit (base 10). Write the following equation using the proper digits.

## ALFA + BETA + GAMA = DELTA

Provide your solution in the comments section below.

Each letter represents a digit (base 10). Write the following equation using the proper digits.

## ALFA + BETA + GAMA = DELTA

Provide your solution in the comments section below.

A simple magic square puzzle this week. What goes in the square to replace the question mark? Post your answer in the comments section below.

My answer is **63**. All the visible numbers are multiples of **7** in sequence. **63** is next in the sequence.

Using four colors, color all regions in the above diagram. Adjacent (sharing a common border) cannot have the same color. Record your answer in the comments section below.

At first glance these appeared to be difficult. They are from a page of about six, of varying difficulty. Anyhow, given four colors to work with the logical approach was to work from the inside circle, using two colors, and then use the remaining two colors for the next ring out and so on. Here is my solution.

Limit yourself to four colors. Color each region so no border shares the same color. It is all right if the same color touches at a corner. If this cannot be done, then explain why. Post your answer in the comments section below.

Here is the result of my effort.

I have not worked this one out, but it should be straight forward to solve I copied and pasted the image from a puzzle site that gave little clarification. Assume anything that appears to be an equilateral triangle is. Post your solution in the comments section below.

Objects in the drawing above that look like squares are squares. The area of the square in the lower left is 5. What is the area of the orange triangle?

Post your answer in the comments section below

This problem becomes easy to solve once you put it into perspective. See the diagram.

Realize the base of the orange triangle is the diagonal of the small square. Then draw the diagonal of the large square. It is parallel to the base of the triangle. The height of the triangle is the diagonal of the medium square, whose area is for times that of the small square. The remainder is left to the reader.

Elmo provided a solution a few days ago. Compare your answer with his.

This week in Abusing Science I want to clue you to an amazing video. Follow the YouTube link.

## MUST WATCH!!! THE EVIDENCE FOR GOD FOUND IN MATHEMATICS!!!

Nov 15, 2019Supernatural By Design

Watch it. The graphics are outstanding.

Of course, the way these things are done is the company in charge of production goes to stock image sources and pays to use the images and the video sequences. Even so…

Yes, this one is all about how numbers demonstrate the proof of God. Isn’t that amazing? And the number 11 (eleven) is special. How special? You will see.

Stunning images.

Look. When you increment by 11, starting at 0, you (after 0) get numbers with repeating digits. Up to 99, that is. After that, well that’s something apparently not in the Bible.

In the image of God. Numbers explain everything.

And here it is. You will be struck dumb at this amazing property of the number 11. When you put up a mirror image of 11, it’s still **11**!

Oh, Jesus. That is truly amazing.

Amazing unless you know a small bit of mathematics, and you know that this only works for numbers to the base 10. Then it becomes apparent this is not merely an abuse of science. It is an insult to the human intellect.

But this is religion, not reality, which makes it clear what we are doing here is peering into the Heart of Dumbness.

I copied this from an Internet site, which I will not identify. I pasted the image full size to be sure there is no problem visualizing it. What fraction of each circle is shaded?

Post your answer in the comments section below. No fair going to the Internet for answers.

Abuse of science is not limited to the unwashed, and political abuse is not a modern invention. In 1897 a proposed Indiana bill would have set the value for π (pi) to three (3). Presumably that would make engineering computations easier in those days without hand-held calculators. It would likely, if put to use, have also resulted in a number of engineering disasters. But we don’t do that crazy shit anymore.

The Ohio state House of Representatives has passed the Student Religious Liberties Act, which prevents teachers from penalizing students for giving incorrect answers on tests or other schoolwork if those facts would conflict with their religious beliefs.

The relevant section reads “No school district board of education (…) shall prohibit a student from engaging in religious expression in the completion of homework, artwork, or other written or oral assignments. Assignment grades and scores shall be calculated using ordinary academic standards of substance and relevance, including any legitimate pedagogical concerns, and shall not penalize or reward a student based on the religious content of a student’s work.”

Ordinarily I would have confronted such inanity head-on and screamed my objections from the roof-tops. But then I had a second thought. This would have come in handy 55 years ago when I was taking a course in differential equations. Disregard that I aced DE, this stuff absolutely ruined my life for several miserable weeks. What if…

What if the Texas legislature, just down the street, had been aware of my misery and decided to come to my aid? I still recall Professor Walston reminding us: “Sometimes a differential equation is solved by staring at it until a solution comes to mind.” Yeah, I would liked to have been able to say, “Professor, you can take this differential equation and put it where the sun don’t shine, because my personal god, the Flying Spaghetti Monster, says the answer is always y = 5x + 3.”

You can see there are times a friendly bit of legislation can be a life saver. And may Jesus have mercy on our souls.

A machine produces parts that are either good (90%), slightly defective (2%), or obviously defective (8%). Produced parts get passed through an automatic inspection machine, which is able to detect any part that is obviously defective and discard it. What is the quality of the parts that make it through the inspection machine and get shipped?

In other words, what is the fraction of good parts in a shipment?

Post your answer in the comments section below.

There are two treasure chests, both closed. One chest contains 100 gold coins. The other contains 50 gold coins and 50 silver coins.

You randomly chose a chest to open, and you do not look inside, but you remove a coin and close the lid. You look at the coin in your hand, and it is gold. What is the probability you chose the chest with 100 gold coins?

Post your answer in the comments section below.

This is a problem in conditional probabilities. What is the probability of A if B is true? It’s written this way.

The probability of A given B is the probability of B given A times the probability of A and divided by the probability of B

A is you chose the chest with 100 gold coins.

B is you pulled a gold coin out of one of the chests.

From the get-go the probability of pulling a gold coin is 3/4.

So P(B) = 1/4

P(B | A) is 1. If you choose the 100% gold chest you are 100% likely to draw a gold coin.

So the answer is ½/¾ = 2/3.

The star measures 5 across. What is the area of the star.

Post your solution in the comments section below.

Because they can’t even.

She’s definitely plotting something.

Because they’ll never meet.

Not unless you Count Dracula.

Because they’re never right.

Use acute angle.

He’ll stop at nothing to avoid them.

They tend to just lose some of their functions.

She’s a perfect 10, but purely imaginary.

Just huddle in the corner, where it’s always 90 degrees.

Probably.

A la mode. Anything else is mean.

But when he rounded them up, he had 300.

**This page has a total of 56 of these, that is if numbers matter.**

Shown here is a cube that fits snugly inside a sphere. It may not appear so from this diagram, but all eight corners of the cube touch the surface of the sphere. The length of each edge of the cube is 8. What is the diameter of the sphere?

It may not appear so, but the solution is ridiculously easy. Post your solution in the comments section below.

This looks hard until you remember the Pythagorean Theorem applies to spaces of all dimensions. I came upon this when a problem at work required computing the distance between two points in a 10-dimensional space.

The diagonal of a rectangle is the square root of the sum of the squares of the two sides. The diagonal of a prism is the square root of the sum of the squares of the three dimensions.

Since the diagonal of the cube is the diameter of the circle, the diameter is 8√3.

Nine circles are circumscribed by a larger circle. The diameter of the small circles is 1. What is the diameter of the large circle?

Post your answer in the comments section below.

The diameter of the small circles is 2. What is the diameter of the large circle.

Post your answer in the comments section below.

See the diagram below.

The centers of the three small circles are at the apexes of equilateral triangle **ABC**. All that is necessary is to determine the distance from **C** to the center of the triangle. Since the sides of the triangle are 2, then the distance to the center is two thirds of 2(½√3) ≅ 1.155. So the radius of the large circle is 1 + 1.155 approximately, and the diameter is 4.309 approximately.

See the two diagrams above. The three large spheres have a diameter of 3. The small sphere has a diameter of 1. The small sphere is stacked in the gap between the three large spheres. What is the value of **h**?

Post your answer in the comments section below.

See the following diagrams.

The centers of the spheres are at the vertices of a quadrahedron whose base is an equilateral triangle. We have **AB** = **BC** = **CA** = 3, and **AD** = **BD** = **CD** = 2. What is the height of the quadrahedron?

The base of the tetrahedron is an equilateral triangle with each side being 3. The altitude of this triangle is 3(√3)/2. The height of the tetrahedron is the altitude of a right triangle with hypotenuse = 2 and base = (2/3) the altitude of the equilateral triangle. So the height of the tetrahedron is 1.

So **h** = 1 + radius of large sphere + radius of small sphere = 3. It’s early in the morning, so you may want to check my math.

The diagrams above are an overhead view and a front view of two spheres backed into a corner. The drawings are not to scale, so believe me when I tell you the radius of the small sphere is 1, and the radius of the large sphere is 4. What is the value of **h**?

Post your answer in the comments section below.

See the following two diagrams.

The radius of the small sphere is 1, so d = √2. Similarly for the large sphere, radius = 3.

We get b = 3√2 – √2 = 2√2.

r = 4

h = 3 + 1 + √(r² – b²) = 4 + √(16 -8) = 4 + √8 ≅ 6.83.

Prove that fractions such as 8/7 must devolve into a repeating decimal sequence.

The answer I gave when I was taking a math class about 1968 was much like Mike’s (see the comments section). It goes like this:

If you divide one integer by another there is either a remainder, or there is not a remainder (zero remainder). The remainder is always going to be less than the divisor, so if you continue repeating the process you will eventually produce a prior remainder. At this point the process will begin to repeat, and the result will be an infinite, repeating sequence.

No fair invoking Kolmogorov complexity. KC stipulates that the result of a process cannot require more information to describe than would be involved in describing the process. Remember, a process for computing the value of π requires only a few words to describe, and the result is a series of digits with no infinitely repeating pattern.

The difference between my answer and Mike’s is mine does not restrict the process to decimal numbers.

The radius of the small circle is 1. What is the radius of the large circles (all the same)?

Post your answer in the comments section below.

This has received several responses, so it’s time to post a solution. See the following diagram.

**R** is the radius of the large circles. 1 is the radius of the small circle.

**8R² = (2 + 2R)² = 4 + 8R + 4R²**

**4R² – 8R – 4 = 0**

**R = 1 + √2**

Mike was the first with the correct answer.