1453 Happy Birthday, Jo Morgan

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Jo Morgan is an inspiring mathematics teacher, collaborator, tweeter, blogger, podcast guest, and author. Today is her birthday!

Jo recently published her first book, A Compendium of Mathematical Methods, and it is all the rage on twitter. Here is a small sampling of tweets expressing excitement for her book:

I can hardly wait until February 4th when Amazon makes it available in the United States!

Jo has enjoyed solving some of my puzzles, so to commemorate her birthday, I’ve made one especially for her. To solve this puzzle, write the numbers 1 to 10 in each of the four sections outlined in bold so that those numbers are the factors of the product clues given in each of the four mixed-up multiplication tables that make up the puzzle. Use logic to solve the puzzle, but I’m warning you, it won’t be easy.

Happy birthday, Jo! I hope you enjoy the puzzle!

Print the puzzles or type the solution in this excel file: 12 Factors 1443-1453

It is convenient for puzzles to be numbered, and this puzzle number is 1453. Here are a few facts about that number:

  • 1453 is a prime number.
  • Prime factorization: 1453 is prime.
  • 1453 has no exponents greater than 1 in its prime factorization, so √1453 cannot be simplified.
  • The exponent in the prime factorization is 1. Adding one to that exponent we get (1 + 1) = 2. Therefore 1453 has exactly 2 factors.
  • The factors of 1453 are outlined with their factor pair partners in the graphic below.

How do we know that 1453 is a prime number? If 1453 were not a prime number, then it would be divisible by at least one prime number less than or equal to √1453. Since 1453 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 31, or 37, we know that 1453 is a prime number.

1453 is the sum of two squares:
38² + 3³ = 1453

That means 1453 is the hypotenuse of a Pythagorean triple:
228-1435-1453 calculated from 2(38)(3), 38² – 3³, 38² + 3³.

Here’s another way we know that 1453 is a prime number: Since its last two digits divided by 4 leave a remainder of 1, and 38² + 3² = 1453 with 38 and 3 having no common prime factors, 1453 will be prime unless it is divisible by a prime number Pythagorean triple hypotenuse less than or equal to √1453. Since 1453 is not divisible by 5, 13, 17, 29, or 37, we know that 1453 is a prime number.

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Countdown to 2020

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It seems that every New Year’s Eve, mathematicians come up with equations with the new year in it. Some of those equations will be a countdown. Here is the equation that I found and made into a gif:

Countdown to 2020

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What will the factors be in the year 2020? I’m here ready with my predictions and the reasons that you can rely on them:
  • 2020 is a composite number.
  • Prime factorization: 2020 = 2 × 2 × 5 × 101, which can be written 2020 = 2² × 5 × 101
  • 2020 has at least one exponent greater than 1 in its prime factorization so √2020 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √2020 = (√4)(√505) = 2√505
  • The exponents in the prime factorization are 2, 1, and 1. Adding one to each exponent and multiplying we get (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12. Therefore 2020 has exactly 12 factors.
  • The factors of 2020 are outlined with their factor pair partners in the graphic below.
A 2020 factor tree may interest you:
Or a factor cake:

 

Here are some other facts about the year 2020:

2020 is palindrome 4C4 in BASE 21 (C is 12 in base 10):
4(21²) + 12(21) + 4(1) = 4(441) + 12(21) + 4 = 1764 + 252 + 4 = 2020

42² + 16² = 2020
38² + 24² = 2020

2020 is the hypotenuse of FOUR Pythagorean triples:
400-1980-2020 which is 20 times (20-99-101)
868-1824-2020 calculated from 38² – 24², 2(38)(24), 38² + 24²
1212-1616-2020 which is (3-4-5) times 404
1344-1508-2020 calculated from 2(42)(16), 42² – 16², 42² + 16²

2020 is the difference of two squares two different ways:
506² – 504² = 2020
106² – 96² = 2020
Can you calculate when 2020 is a leg in a Pythagorean triple from those equations? (The Pythagorean triples can be calculated from a² – b², 2(a)(b),  a² + b²)

Then can you calculate other times 2020 is a leg in a Pythagorean triple from these facts? (The Pythagorean triples can be calculated from 2(a)(b), a² – b²,  a² + b²)
2(1010)(1) = 2020
2(505)(2) = 2020
2(202)(5) = 2020
2(101)(10) = 2020
Did any of those equations produce the same Pythagorean triples that the difference of two squares produced?

2020 is the sum of four squares in MANY different ways. Here is how I found two of those ways:

I will end this post by sharing mathematical facts about the number 2020 that I’ve seen on twitter including some of my own tweets.

 

 

 

 

That’s a lot of love for the number 2020. Have a wonderful year, everybody!

Let’s Make a Factor Cake for 2020

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We often celebrate special occasions with a cake!

Coincidentally, there is a method to find the prime factorization of a number that is called the cake method.

Let’s make a factor cake for the year 2020 to celebrate its arrival!
2020 Factor Cake

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The factor cake shows that the prime factorization of 2020 is 2 × 2 × 5 × 101. We can write that more compactly: 2020 = 2² × 5 × 101.
In case you would like a still picture of the cake instead of the gif, here it is:
I will write more about the number 2020 before tomorrow. Enjoy saying good-bye to 2019 and getting ready for the new year!

1452 Poinsettia Plant Mystery

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Merry Christmas, Everybody!

The poinsettia plant has a reputation for being poisonous, but it has never been a part of a whodunnit, and it never will. Poinsettias actually aren’t poisonous.

Multiplication tables might also have a reputation for being deadly, but they aren’t either, except maybe this one. Can you use logic to solve this puzzle without it killing you?

To solve the puzzle, you will need some multiplication facts that you probably DON’T have memorized. They can be found in the table below. Be careful! The more often a clue appears, the more trouble it can be:

Notice that the number 60 appears EIGHT times in that table. Lucky for you, it doesn’t appear even once in today’s puzzle!

Now I’d like to factor the puzzle number, 1452. Here are a few facts about that number:

1 + 4 + 5 + 2 = 12, which is divisible by 3, so 1452 is divisible by 3.
1 – 4 + 5 – 2 = 0, which is divisible by 11, so 1452 is divisible by 11.

  • 1452 is a composite number.
  • Prime factorization: 1452 = 2 × 2 × 3 × 11 × 11, which can be written 1452 = 2² × 3 × 11²
  • 1452 has at least one exponent greater than 1 in its prime factorization so √1452 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1452 = (√484)(√3) = 22√3
  • The exponents in the prime factorization are 2, 1, and 2. Adding one to each exponent and multiplying we get (2 + 1)(1 + 1)(2 + 1) = 3 × 2 × 3 = 18. Therefore 1452 has exactly 18 factors.
  • The factors of 1452 are outlined with their factor pair partners in the graphic below.

To commemorate the season, here’s a factor tree for 1452:

Have a very happy holiday!

1451 Star of Wonder

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An important part of Mathematics is noticing patterns. I love it when mathematicians ask students, “What do you notice? What do you wonder?”

Those are questions you can ponder as you gaze on this star of wonder made from several different graphs.

 

To help distinguish the graphs, the dotted lines are exponential functions, the dashed lines are natural logarithm functions, and the solid lines are linear functions.

What do you notice? What do you wonder?

Now I’ll tell you a little bit about the post number, 1451:

  • 1451 is a prime number.
  • Prime factorization: 1451 is prime.
  • 1451 has no exponents greater than 1 in its prime factorization, so √1451 cannot be simplified.
  • The exponent in the prime factorization is 1. Adding one to that exponent we get (1 + 1) = 2. Therefore 1451 has exactly 2 factors.
  • The factors of 1451 are outlined with their factor pair partners in the graphic below.

How do we know that 1451 is a prime number? If 1451 were not a prime number, then it would be divisible by at least one prime number less than or equal to √1451. Since 1451 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 31, or 37, we know that 1451 is a prime number.

1450 A Pair of Factor Trees

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On today’s puzzle, there are two small Christmas trees. Will two smaller trees on the puzzle be easier to solve than one big one? You’ll have to try it to know!

Every puzzle has a puzzle number to distinguish it from the others. Here are some facts about this puzzle number, 1450:

  • 1450 is a composite number.
  • Prime factorization: 1450 = 2 × 5 × 5 × 29, which can be written 1450 = 2 × 5² × 29
  • 1450 has at least one exponent greater than 1 in its prime factorization so √1450 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1450 = (√25)(√58) = 5√58
  • The exponents in the prime factorization are 1, 2, and 1. Adding one to each exponent and multiplying we get (1 + 1)(2 + 1)(1 + 1) = 2 × 3 × 2 = 12. Therefore 1450 has exactly 12 factors.
  • The factors of 1450 are outlined with their factor pair partners in the graphic below.

In case you are looking for factor trees for 1450, here are two different ones:

1450 is the hypotenuse of SEVEN Pythagorean triples:
170-1440-1450 which is 10 times (17-144-145)
240-1430-1450 which is 10 times (24-143-145)
406-1392-1450 which is (7-24-25) times 58
666-1288-1450 which is 2 times (333-644-725)
728-1254-1450 which is 2 times (364-627-725)
870-1160-1450 which is (3-4-5) times 290
1000-1050-1450 which is (20-21-29) times 50

1449 Christmas Star

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If you’ve ever wished you knew the multiplication table better, then make that wish upon this Christmas star. If you use logic and don’t give up,  then you can watch your wish come true!

I number the puzzles to distinguish them from one another. That star puzzle is way too big for a factor tree made with its puzzle number:

Here’s more about the number 1449:

  • Prime factorization: 1449 = 3 × 3 × 7 × 23, which can be written 1449 = 3² × 7 × 23
  • 1449 has at least one exponent greater than 1 in its prime factorization so √1449 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1449 = (√9)(√161) = 3√161
  • The exponents in the prime factorization are 2, 1, and 1. Adding one to each exponent and multiplying we get (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12. Therefore 1449 has exactly 12 factors.
  • The factors of 1449 are outlined with their factor pair partners in the graphic below.

1449 is the difference of two squares in 6 different ways:
725² – 724² = 1449
243² – 240² = 1449
107²-100² = 1449
85² – 76² = 1449
45² – 24² = 1449
43² – 20² = 1449

1448 Christmas Factor Tree

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Here’s a puzzle that looks a little like a Christmas tree. Some of the clues might give you a little bit of trouble. For example, the common factor of 60 and 30 might be 5, 6, or 10. Likewise, the common factor of 8 and 4 might be 1, 2, or 4.

Which factor should you use? Look at all the other clues and use logic. Logic can help you write each of the numbers 1 to 12 in both the first column and the top row so that the given clues and those numbers behave like a multiplication table. Good luck!

I have to number every puzzle. It won’t help you solve the puzzle, but here are some facts about the number 1448:

The number made by its last two digits, 48, is divisible by 4, so 1448 is also divisible by 4. That fact can give us the first couple of branches of 1448’s factor tree:

  • 1448 is a composite number.
  • Prime factorization: 1448 = 2 × 2 × 2 × 181, which can be written 1448 = 2³ × 181
  • 1448 has at least one exponent greater than 1 in its prime factorization so √1448 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1448 = (√4)(√362) = 2√362
  • The exponents in the prime factorization are 3 and 1. Adding one to each exponent and multiplying we get (3 + 1)(1 + 1) = 4 × 2 = 8. Therefore 1448 has exactly 8 factors.
  • The factors of 1448 are outlined with their factor pair partners in the graphic below.

1448 is also the hypotenuse of a Pythagorean triple:
152-1440-1448 which is 8 times (19-180-181)

1447 Christmas Light Puzzle

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If you’ve ever had a string of lights go out because ONE bulb went bad, it can be a very frustrating puzzle to figure out which light is causing the problem.

This is not that kind of puzzle. For this one, you just need to figure out where to put the numbers from 1 to 12 in both the first column and the top row so that the given clues are the products of those numbers. There is only one solution, and if you always use logic, it will not be a frustrating puzzle to solve.

I gave that puzzle the puzzle number 1447. That number won’t help you solve the puzzle, but here are some facts about it anyway:

  • 1447 is a prime number.
  • Prime factorization: 1447 is prime.
  • 1447 has no exponents greater than 1 in its prime factorization, so √1447 cannot be simplified.
  • The exponent in the prime factorization is 1. Adding one to that exponent we get (1 + 1) = 2. Therefore 1447 has exactly 2 factors.
  • The factors of 1447 are outlined with their factor pair partners in the graphic below.

How do we know that 1447 is a prime number? If 1447 were not a prime number, then it would be divisible by at least one prime number less than or equal to √1447. Since 1447 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, or 37, we know that 1447 is a prime number.

1447 is also the difference of two consecutive squares:
724² – 723² = 1447

 

 

 

1446 Peppermint Stick

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Red and green striped peppermint sticks are often seen in stores and homes in December. Can you lick this peppermint stick puzzle or will you let it lick you?

The puzzle number was 1446. Here are a few facts about that number:

  • 1446 is a composite number.
  • Prime factorization: 1446 = 2 × 3 × 241
  • 1446 has no exponents greater than 1 in its prime factorization, so √1446 cannot be simplified.
  • The exponents in the prime factorization are 1, 1, and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 = 8. Therefore 1446 has exactly 8 factors.
  • The factors of 1446 are outlined with their factor pair partners in the graphic below.

1446 is also the hypotenuse of a Pythagorean triple:
720-1254-1446 which is 6 times (120-209-241)