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This is the Transum Newsletter for the month of August 2024 and whether you are on holiday or not, here is the puzzle of the month for your consideration:

I am thinking of three numbers.

If I double them all the mean stays the same.

If I square all the numbers the new median is nine.

The range of the numbers is seven.

The square of the product of the numbers is 144.

What are the three numbers?

If you get an answer I'd love to hear how you solved it (or your students solved it). Drop me an email at: gro.musnarT@rettelsweN

While you think about that here are some of the key resources added to the Transum website during the last month.

Fraction Foundations is brand new. It is the most basic of all the fraction activities on the website and fits in nicely with the Functional Skills curriculum.

A new help video has been made for the Logarithms exercise.

There's a brand-new Animal Records matching activity. This interactive exercise is designed to engage and challenge learners as they match animal records with their corresponding values and units. From the fastest land animal to the loudest creature in the ocean, this activity covers a diverse range of animal world records and requires estimating skills and a knowledge of common units of measure. Give it a try yourself and see how many records you can match correctly!

Using the same matching-three-things format I then created the Domain, Range and Asymptotes matching activity in response to a subscriber's request for resources on domains and ranges. While Transum might not offer the same variety of questions as a textbook due to the limitations of the input format for automatic marking, this new activity adds some variety. It could also work well as a pairs activity, providing an engaging alternative to traditional textbook exercises.

Finally, to make the most of the matching activity software I had developed, I created a third activity focused on numeracy. It is called Products of Primes and will make a good challenge for those students learning the divisibility tests.

Are your students on their long summer holiday at the moment (sorry southern hemisphere)? If so recreational Maths activities not only reinforce academic concepts but also promote critical thinking, creativity, and a love for learning. So here's an idea, send your students an email containing a link to my collection of Holiday Maths Activities.

Don't forget that there is an Olympic Puzzle and ideas for investigations to work through in between watching the track and field events.

The book I am really enjoying at the moment is A Year in Numbers by Kyle D Evans. The book contains a fact for every day of the year explained in a way that makes the facts easy to understand and very interesting. Do you know what fraction of the numbers one to a hundred are prime? I had never considered it before and was surprised to find out that it is exactly one quarter. That's an interesting question that students can be asked on completing the Sieve of Eratosthenes.

I might never teach alternate angles again without telling the story of the calculation that so clearly and spectacularly made use of them:

Eratosthenes measured Earth’s circumference around 240 BC by comparing shadows. In Syene (modern Aswan), a well had no shadow at noon on the summer solstice, indicating the Sun was directly overhead. In Alexandria, 800 km north, a vertical stick cast a shadow, showing the Sun’s rays hit at an angle of about 7.2 degrees.

He reasoned that this angle corresponded to the arc between Alexandria and Syene, which is 1/50th of a circle (7.2 degrees is 1/50th of 360 degrees). Thus, the Earth’s circumference is 50 times the distance between the two cities, approximately 40,000 km.

cmglee, David Monniaux, jimht at shaw dot ca, CC BY-SA 4.0, via Wikimedia Commons

Finally the answer to last month's puzzle which was:

Pirate Pearl sits in a hidden cove counting her treasure of 50 dazzling precious stones of three distinct colours. There are more fiery red stones than cool blue ones, and more blue stones than verdant green ones. If you were to pluck 3 stones at random, the chance of picking one of each colour is exactly one-tenth. Can you figure out how many stones of each colour there are?

The answer is 35 red stones, 8 blue stones and 7 green stones.

The first five correct answers were received from Mala, Leonard, Kevin, Farquad and Rick. You can see some of the methods below.

That's all for now,

John

P.S. I was very proud of the mathematician who entered the Olympic weightlifting event. He excelled at lifting the average!

Do you have any comments? It is always useful to receive feedback on this newsletter and the resources on this website so that they can be made even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.

Mala, New Zealand

Monday, July 1, 2024

"The number of precious stones of each colour are as follows: 35 fiery red, 8 cool blue and 7 verdant green.

I used a mix of probability and algebraic skills to solve the problem.

Here it goes:

The number of stones in each colour multiply to 1960 (using probability methods) and the only factors of 1960 that add to 50 are 35, 8 and 7."

Leonard, US

Tuesday, July 2, 2024

"I was listening to the July puzzle this morning, and decided to give it a try.

After a couple of false starts, I came up with the following thinking.

- Probability of a single draw = \(R/50 * B/49 * G/48\).

- With 6 ways to arrange the drawn R/B/G gems, we want \(6(R/50 * B/49 * G/48) = 0.1\), where 0.1 is our 10% goal.

- Simplifying yields \(R * B * G = 50 * 49 * 48 / 60 = 117,600 / 60 = 1,960\).

- Prime factoring 1,960 yields \(2*2*2*5*7*7\), leading to possibly values of G of 2, 4, 5, 7, 8, 10, and 14.

- Working my way up from 2 (utilizing the constraint R+B+G = 50) led me to my answer of 35, 8 and 7.

I hope that is correct.

Love the website."

Farquad,

Friday, July 5, 2024

"Let’s say there are x red, y blue and z green, such that x+y+z = 50

The info we are given lets us deduce that the number of total outcomes is 50C3, or 19600.

The number of favourable outcomes would be the number of ways you can get a red, * the number of ways to get a blue * the number of ways to get a green, or xyz

so our total probability is xyz/19600 = 1/10, giving us xyz = 1960

combining this with x+y+z = 50 lets us find that the only solution that satisfies both equations, and x > y > z is:

x = 35

y = 8

z = 7"

Rick, Regular Podcast Listener

Monday, July 8, 2024

"There are six ways to pull three unique stones from the collection, where R=red, G=green, and B=blue: RGB, RBG, GRB, GBR, BRG, and BGR.

The probability of pulling the first combination is R/50 x G/49 x B/48. This is the same for all the other combinations as well, only the order of the numerator changes.

Adding the six together and equating to the stated probability results in 6 x RGB / (50 x 49 x 48) = 1/10.

Solving for RGB and distributing the 10 and 6 appropriately = 50 / 10 x 49 x 48 / 6 = 5 x 7 x 7 x 2 x 2 x 2.

Since R + G + B = 50, an even number, and there are three odd and even factors, either all the odd factors need to be multiplied by an even factor (which would result in two values being the same, contradicting the statement that red was greater than blue which was greater than green). Therefor, assume one of the values is 35, leaving 7 and 8. Eureka. Red = 35, blue = 8, and green = 7, which, when added together, total 50. "

Kirby,

Tuesday, July 9, 2024

"35 Red, 8 Blue, 7 Green.If you pick one at a time there are 6 ways to get one of each - RBG, RGB, BRG, BGR, GRB, GBR. The odds of any one of these scenarios is RxBxG/(50x49x48). So the odds you hit one of these scenarios is 6xRxBxG/(50x49x48) = 1/10. Simplifying you get RxBxG = 1960. You also know that R + B + G = 50. From there I looked up the prime factors of 1960 and tinkered with combining them until I found 3 that added up to 50. Assigned the order based on the questions saying more reds than blues and more blues than greens."

Alfonso Rodríguez, Mexico

Monday, August 5, 2024

"Amazing explanation about the earth´s curvature. I had never really understood this calculation until I followed this explanation."