Christmas

Eight Desmos Ornaments

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I made some Christmas ornaments in Desmos that I hope you will enjoy. If you click on and off the circles on the left of the descriptions, you can see all eight ornaments in one Desmos graph, or you can find them all pictured below in this post. If you click the arrow next to each description in Desmos, you can also see the equations used to produce each ornament. However, the snowflake and Rudolf’s face required many ordered pairs, which I put into a separate folder.

1. Decorated half red and half green:

2. Decorated with diagonal stripes:

3. Decorated with sines and secants:

4. Decorated with a snowflake:

5. Decorated with a checkerboard design: (This was a pleasant surprise that required only one equation!)

6. Decorated with a spiral

7. Decorated with ellipses for a 3D look:

8. Decorated with Rudolf’s face:

Perhaps you will choose to make an ornament yourself in Desmos. If so, I’d love to see it.

I hope you all have a very merry Christmas!

1771 Pascal’s Triangle and the Twelve Days of Christmas

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A Twelve Days of Christmas Puzzle with Triangular and Tetrahedral Numbers:

I wanted a copy of Pascal’s triangle with 14 rows. I couldn’t find one, so I made my own. To fill in the missing number in a cell, simply write the sum of the two numbers above it. I would suggest filling it in together as a class so that they can see how it is done without actually having to write in all the numbers themselves. The biggest missing sum is 364.

After filling that puzzle in together as a class, I would give students this next copy of Pascal’s triangle to use.

There are many patterns in Pascal’s triangle. It can be fun to color them with that in mind. I would caution students to color lightly so that they can still read the numbers afterward. How did I color this one? If the number in a cell is not divisible by the row number, I colored it green. Of course, all the 1’s were colored green. If all the other numbers in the row were divisible by the row number, I colored all of them red. If only some of them were, I colored them yellow. Notice that the row number of every row that is red is a prime number. Composite row numbers will always have at least one entry that is not divisible by the row number.

I divided each of the numbers in this next one by 3, noted the remainder, and colored them accordingly:

  • remainder 0 – red
  • remainder 1 – green
  • remainder 2 – yellow

1771 is a Tetrahedral Number:

364 = 12·13·14/6. That means it is the 12th tetrahedral number.
If my true love gave me all the gifts listed in the Twelve Days of Christmas song, it would be a total of 364 gifts. Since I don’t have use for all those birds, if I returned one gift a day, it would take me 364 days to return them all. That’s one day less than an entire year!

1771 = 21·22·23/6. That means it is the 21st tetrahedral number.
If there were 21 days of Christmas, and the pattern given in the song held, my true love would give me 1771 gifts. Yikes, I’ll need a bigger house or maybe a bird sanctuary!

Here is one-half of the 23rd row in Pascal’s triangle showing the number 1771:

I’ve also been thinking about the next tetrahedral number after 1771 because the year, 2024, has almost arrived. Note to my true love: I don’t need or want 2024 gifts, please!

Factors of 1771:

1771 is a palindrome with an even number of digits, so 1771 is divisible by eleven.

  • 1771 is a composite number.
  • Prime factorization: 1771 = 7 × 11 × 23.
  • 1771 has no exponents greater than 1 in its prime factorization, so √1771 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 1771 has exactly 8 factors.
  • The factors of 1771 are outlined with their factor pair partners in the graphic below.

More About the Number 1771:

1771 is the difference of two squares in four ways:

886² – 885² = 1771,
130² – 123² = 1771,
86² – 75² = 1771, and
50² – 27² = 1771.

It is easy to see that 1771 is a palindrome in base 10, but it is also a palindrome in some other bases:
It’s 4H4 in base 19 because 4(19²)+17(19)+4(1)=1771,
232 in base 29 because 2(29²)+3(29)+2(1)
1T1 in base 30 because 1(30²)+29(30)+1(1)=1771, and
NN in base 76 because 23(76)+23(1).

1769 I Want a Hip Hypotenuse for Christmaths

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Today’s Puzzle:

Why do I want a hip hypotenuse for Christmaths?

Many have heard the equation a² + b² = c² to help find the hypotenuse of a right triangle when given two legs. What do you do if you are given the hypotenuse and a leg instead of two legs? You use b² = c² – a².

Sometimes finding squares and taking square roots is not too difficult:

  • x = √(29²-21²)                              x = √(61² – 11²)
  • x = √(841-441)                           x = √(3721 – 121)
  • x = √400                                         x = √3600
  • x = 20                                              x = 60

Other times it can be more challenging:

  • x = √(177² – 48²)
  • x = √(31329 – 2304)
  • x = √29025
  • x ≈ 170.37, that’s irrational and not in the simplest form. Finding the factors of 29025, so its square root can be simplified, is going to be a pain!

Try this instead:

  • x = √(177² – 48²) That’s the difference of two squares, so it can be factored!
  • x = √((177 – 48)(177 + 48))
  • x = √(129 · 225) I love that I have two factors instead of one big number! And in this case, one of them is a perfect square! 225 = 15².
  • x = √(3 · 43 · 15²)
  • x = 15√129.

Most people learn the Pythagorean theorem before they learn how to factor the difference of two squares, but then they never apply it to the Pythagorean theorem. Once you know both concepts, factor whenever you can!

A Math Parody of the Song, “I Want a Hippopotamus for Christmas”

On Thanksgiving, my son gave me an early Christmas present: a t-shirt that had a right triangle with a hippopotamus sprawled over the hypotenuse. The shirt had the words, “I want a hippopotenus for Christmath” at the bottom.

Somebody suggested that I sing it.

I looked for a mathematical version of I Want a Hippopotamus for Christmas and couldn’t find one, so I made my own. I decided I liked “hip hypotenuse” better than “hippopotenuse” and that the British “maths” sounded better than the American “math.” Then I recorded it. If I had more time, I would have waited until I didn’t have a cold and would have worked on my timing a bit more. Since I wanted it to be ready to present on a certain day at school, I just went with it as is. I hope you enjoy it.

I shared it with my family. Here’s how one of my sons responded:

Well, some moms have made negative comments about their kid’s ability to learn some math concepts. Indeed, I didn’t.

I already had been thinking about reworking some of the lyrics, making the first half of the song about the Pythagorean theorem and the second half about Trigonometry. I’ll re-record it sometime, but here are the lyrics I’ll use for the revised version:

I want a hip hypotenuse for Christmaths.
Only a hip hypotenuse will do.
I don’t want a doll, no dinky tinker toy;
I want a hip hypotenuse to play with and enjoy!

I want a hip hypotenuse for Christmaths.
I don’t think Santa Claus will mind, do you?
He won’t have to use those squares to find the last side, too.
Just the difference times the sum. That’s an easy root to do.

I can see me now on Christmaths morning calculating squares.
Oh, what joy and what surprise
When I open up my eyes
To see that hip hypotenuse given there.

I want a hip hypotenuse for Christmaths,
Only a hip hypotenuse will do.
No crocodiles, eating more or lesses.
I only like hip hypotenuses.
And hip hypotenuses like me, too.

I want a hip hypotenuse for Christmaths.
A hip hypotenuse is all I want.
Mom says triangles are often right, for them
Teacher taught a theorem that is Pythagorean.

I want a hip hypotenuse for Christmaths,
The kind that’s used in trigonometry.
The sine of an angle is its opposite side
Over the hip hypotenuse. Make sure it’s simplified!

I can see me now on Christmaths solving triangles downstairs.
With the laws of sines and cosines.
Each time(?!) I must choose
When there’s no hypotenuse anywhere.

I WANT a hip hypotenuse for Christmaths,
Only a hip hypotenuse will do.
No crocodiles, eating more or lesses.
I only like hip hypotenuses.
And hip hypotenuses like me, too.

I hope my song made you laugh. For more laughs, check out this Statistics Saturday post at Another Blog, Meanwhile. He lists several humouous hippopotamus-related unwise Christmas gifts.

Factors of 1769:

This is my 1769th post. What are the factors of 1769?

  • 1769 is a composite number.
  • Prime factorization: 1769 = 29 × 61.
  • 1769 has no exponents greater than 1 in its prime factorization, so √1769 cannot be simplified.
  • The exponents in the prime factorization are 1 and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1769 has exactly 4 factors.
  • The factors of 1769 are outlined with their factor pair partners in the graphic below.

More About the Number 1769:

1769 is the sum of two squares in two different ways:

40² + 13² = 1769, and
37² + 20² = 1769.

1769 is the hypotenuse of FOUR Pythagorean triples:
319-1740-1769, which is (11-60-61) times 29,
969-1480-1769, calculated from 37² – 20², 2(37)(20), 37² + 20²,
1040-1431-1769, calculated from 2(40)(13), 40² – 13², 40² + 13²,
1220-1281-1769, which is (20-21-29) times 61.

Did you notice that 20-21-29 and 11-60-61 were the two triangles used in today’s puzzle? It would not have been so easy if I had used 319-1740-1769 and 1220-1281-1769 instead!

1769 is a palindrome in some other bases:
It’s 585 in base 18,
1I1 in base 34, and
TT in base 60.

1768 A Polygonal Christmas Tree on Desmos

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I created this polygonal Christmas tree with a polygonal star on Desmos, and it looks like it is living and breathing to me!

Later I saw this Christmas tree post and decided to share it here:

Today’s Puzzle:

Can you find the factors that belong on this Christmas factor tree for 1768?

Factors of 1768:

  • 1768 is a composite number.
  • Prime factorization: 1768 = 2 × 2 × 2 × 13 × 17, which can be written 1768 = 2³ × 13 × 17.
  • 1768 has at least one exponent greater than 1 in its prime factorization so √1768 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1768 = (√4)(√442) = 2√442.
  • The exponents in the prime factorization are 3,1 and 1. Adding one to each exponent and multiplying we get (3 + 1)(1 + 1) (1 + 1) = 4 × 2 × 2 = 16. Therefore 1768 has exactly 16 factors.
  • The factors of 1768 are outlined with their factor pair partners in the graphic below.

More About the Number 1768:

1768 is the sum of two squares in two different ways:
38² + 18² = 1768, and
42² + 2² = 1768.

1768 is the hypotenuse of FOUR Pythagorean triples:
168-1760-1768, calculated from 2(42)(2), 42² – 2², 42² + 2²,
680-1632-1768, which is 136 times (5-12-13),
832-1560-1768, which is 104 times (8-15-17),
1120-1368-1768, calculated from 38² – 18², 2(38)(18), 38² + 18².

The first triple is also 8 times (21-220-221), and
the last triple is also 8 times (140-171-221).

1768 looks interesting in some bases you probably would never care about:

It’s 404 in base 21 because 4(21²) + 0(21) + 1(1) = 1768.
It’s 1Q1 in base 31,
YY in base 51, and
QQ in base 67.

Can you solve for Q and Y?

1724 Carol of the Ukrainian Bells

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Today’s Puzzle:

As I sit in my warm, peaceful house, I often think about the people of Ukraine whose country has been ravaged by war. Many of them, including children, are also facing a winter with no heat. My heart goes out to them.

My childhood was so unlike theirs. I was able to attend school classes and learn about many different topics without fear of dying. In Junior High School choir class, one of the songs I learned was called Carol of the Bells. I just recently learned from Slate magazine of the song’s Ukrainian roots:

A little over a hundred years ago Mykola Leontovych, a Ukrainian composer, arranged several of his country’s folk songs together in a piece he titled Shchedryk. Tragically, he was murdered by a Russian assassin on January 23, 1921, in the Red Terror, when the Bolsheviks were intent on eliminating Ukrainian leaders, intellectuals, and clergy.

During this time of great unrest, the Ukrainian National Chorus performed Shchedryk around the world and in cities large and small in the United States. One performance was even given at the famed Carnegie Hall on October 5, 1922. The haunting melody was heard by Peter Wilhousky who penned alternate words for it: Hark how the bells, Sweet silver bells,…

Now it is one of our most beloved Christmas carols. I am grateful I learned the words and tune in junior high, although I wish I had learned of its Ukrainian history then as well.

These two bells puzzles are reminiscent of Ukraine’s flag. Long may it wave. Write each number 1 to 10 in the yellow columns and rows so that the given clues are the products of the numbers you write.

Here are the same puzzles if you prefer to use less of your printer ink.

Factors of 1724:

  • 1724 is a composite number.
  • Prime factorization: 1724 = 2 × 2 × 431, which can be written 1724 = 2² × 431.
  • 1724 has at least one exponent greater than 1 in its prime factorization so √1724 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1724 = (√4)(√431) = 2√431.
  • The exponents in the prime factorization are 2 and 1. Adding one to each exponent and multiplying we get (2 + 1)(1 + 1) = 3 × 2 = 6. Therefore 1724 has exactly 6 factors.
  • The factors of 1724 are outlined with their factor pair partners in the graphic below.

More About the Number 1724:

1724 is the difference of two squares:
432² – 430² = 1724.

1724₁₀ = 464₂₀, a palindrome, because
4(20²) + 6(20¹) + 4(20°) = 1724.

1704 Christmas Factor Tree

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Today’s Puzzle:

If you know the factors of the clues in this Christmas tree, and you use logic, it is possible to write each number from 1 to 12 in both the first column and the top row to make a multiplication table. It’s a level six puzzle, so it won’t be easy, even for adults, but can YOU do it?

Factors of 1704:

If you were expecting to see a factor tree for the number 1704, here is one of several possibilities:

  • 1704 is a composite number.
  • Prime factorization: 1704 = 2 × 2 × 2 × 3 × 71, which can be written 1704 = 2³ × 3 × 71.
  • 1704 has at least one exponent greater than 1 in its prime factorization so √1704 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1704 = (√4)(√426) = 2√426.
  • The exponents in the prime factorization are 3,1 and 1. Adding one to each exponent and multiplying we get (3 + 1)(1 + 1) (1 + 1) = 4 × 2 × 2 = 16. Therefore 1704 has exactly 16 factors.
  • The factors of 1704 are outlined with their factor pair partners in the graphic below.


More About the Number 1704:

1704 is the difference of two squares in FOUR different ways:
427² – 425² = 1704,
215² – 211² = 1704,
145² – 139² = 1704, and
77² – 65² = 1704.

Why was Six afraid of Seven? Because Seven ate Nine.
1704 is 789 in a different base:
1704₁₀ = 789₁₅ because 7(15²) + 8(15¹) + 9(15º) = 1704.

1568 A Challenging Christmas Tree

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Today’s Puzzle:

Can you write the numbers from 1 to 10 in each of the boldly outlined columns and rows so that each quadrant of this Christmas tree puzzle behaves like a multiplication table? Remember to use logic and not guessing and checking.

Here is the same puzzle that won’t use so much ink to print:

1568 Factor Tree:

Here is one of several possible factor trees for 1568:

Factors of 1568:

  • 1568 is a composite number.
  • Prime factorization: 1568 = 2 × 2 × 2 × 2 × 2 × 7 × 7, which can be written 1568 = 2⁵ × 7².
  • 1568 has at least one exponent greater than 1 in its prime factorization so √1568 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1568 = (√784)(√2) = 28√2.
  • The exponents in the prime factorization are 5 and 2. Adding one to each exponent and multiplying we get (5 + 1)(2 + 1) = 6 × 3 = 18. Therefore 1568 has exactly 18 factors.
  • The factors of 1568 are outlined with their factor pair partners in the graphic below.

More about the Number 1568:

1568 is the difference of two squares SIX different ways:
393² – 391² = 1568,
198² – 194² = 1568,
102² – 94² = 1568,
63² – 49² = 1568,
57² – 41² = 1568, and
42² – 14² = 1568.

1567 Peppermint Stick

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Today’s Puzzle:

Our mystery level puzzle looks like a sweet stick of Christmas candy. Will solving it be sweet or will it be sticky? You’ll have to try it yourself to know.

Factors of 1567:

  • 1567 is a prime number.
  • Prime factorization: 1567 is prime.
  • 1567 has no exponents greater than 1 in its prime factorization, so √1567 cannot be simplified.
  • The exponent in the prime factorization is 1. Adding one to that exponent we get (1 + 1) = 2. Therefore 1567 has exactly 2 factors.
  • The factors of 1567 are outlined with their factor pair partners in the graphic below.

How do we know that 1567 is a prime number? If 1567 were not a prime number, then it would be divisible by at least one prime number less than or equal to √1567. Since 1567 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, or 37, we know that 1567 is a prime number.

More about the Number 1567:

1567 is the sum of two consecutive numbers:
783 + 784 = 1567.

1567 is also the difference of two consecutive squares:
784² – 783² = 1567.

1565 Stable with Manger

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Today’s Puzzle:

This mystery level puzzle reminds me of the manger in the stable that first Christmas night.

How difficult will the puzzle be to solve? That is part of the mystery. You will have to try it for yourself to find out.

Factors of 1565:

  • 1565 is a composite number.
  • Prime factorization: 1565 = 5 × 313.
  • 1565 has no exponents greater than 1 in its prime factorization, so √1565 cannot be simplified.
  • The exponents in the prime factorization are 1 and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1565 has exactly 4 factors.
  • The factors of 1565 are outlined with their factor pair partners in the graphic below.

More about the Number 1565:

1565 is the sum of two squares in two different ways:
37² + 14² = 1565, and
38² + 11² = 1565.

125-1560-1565, which is 5 times (25-312-313),
836-1323-1565, calculated from 2(38)(11), 38² – 11², 38² + 11²,
939-1252-1565, which is (3-4-5) times 313, and
1036-1173-1565, calculated from 2(37)(14), 37² – 14², 37² + 14².

1563 The Holly Wreath

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Today’s puzzle:

A holly wreath is yet another symbol that connects Christmas with Easter. It symbolizes eternity in its color and shape. It bears white flowers, red berries, and thorns reminding us of purity, blood, and a crown of thorns.

You might find some of the clues in this level 5 puzzle to be like thorns, but don’t give up. Use logic and perseverance and you will be able to find its unique solution.

Here’s the same puzzle without all the added color:

Factors of 1563:

  • 1563 is a composite number.
  • Prime factorization: 1563 = 3 × 521.
  • 1563 has no exponents greater than 1 in its prime factorization, so √1563 cannot be simplified.
  • The exponents in the prime factorization are 1 and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1563 has exactly 4 factors.
  • The factors of 1563 are outlined with their factor pair partners in the graphic below.

Another Fact about the Number 1563:

1563 is the hypotenuse of a Pythagorean triple:
837-1320-1563, which is 3 times (279-440-521).