Month: December 2023

Facts and Factors for the Year 2024

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A Countdown to 2024:

2024 Countdown

make science GIFs like this at MakeaGif
Here are some other countdowns:

What Kind of Factors Will the Year 2024 Bring Us?

Here’s a factor cake to celebrate 2024’s arrival:

And its factor pairs are outlined on this chart:

The sum of all the factors of a number (excluding itself) determines if a number is deficient, perfect, or abundant. Which of those describes 2024?

Powerful Facts About the Number 2024:

2³ + 3³ + 4³ + 5³ + 6³ + 7³ + 8³ + 9³ = 2024, as illustrated below

2¹º + 10³ =2024.

2024 is the sum of eleven consecutive even square numbers:

2024 is the difference of two squares in FOUR different ways:

507² – 505² = 2024,
255² – 251² = 2024,
57² – 35² = 2024, and
45² – 1² = 2024.

One of those four equations brings us to…

Today’s Puzzle and Some Other 2024-Themed Puzzles:

Here are some other 2024-themed puzzles:

2024 in Pythagorean Triple Triangles

2024 is a leg in quite a few Pythagorean triple triangles. Here are a few:

I didn’t even include all the triangles listed on the left in the illustration because some of the points were too close together with the scale I used in Desmos.

I also didn’t include
2024² + 512070² = 512074 or
2024² + 128010² = 128026
because those triangles would have made the scale even worse.

There are more triangles, but I think this is a good enough representation.

Here’s a more complete list:

2024 is NOT the hypotenuse of any Pythagorean triple because none of its prime factors leave a remainder of 1 when divided by 4.

2024 in Pascal’s Triangle:

2024 is in the 24th row of Pascal’s triangle. Click on the image to see the numbers in the image better.

Since it is in the third column of that 24th row, 2024 is a Tetrahedral number. That means it is the sum of the first 22 triangular numbers. That fact can be illustrated as I have here in this vertically rotating Desmos 3D image. Go ahead and click on the image below to see this tetrahedron rotating more horizontally. Also, notice that I made the image with 2024 quarter-unit spheres.

As I stated before, the image is made from the first 22 triangular numbers stacked on top of each other. The sum of the first 22 triangular numbers is given below:

This tetrahedral number can also be expressed mathematically in this way:

This way:

The second way listed here:

Or this way:

2024 Consecutive Number Sums

2024 is the sum of consecutive counting numbers in three different ways:

2024 is the sum of consecutive odd numbers in four different ways: (It’s because 2024 is the difference of two squares in four different ways.)

2024 is the sum of consecutive even numbers:

8 consecutive even numbers:
246+248+250+252+254+256+258+260=2024.

11 consecutive even numbers:
174+176+178+180+182+184+186+188+190
+192+194=2024.

23 consecutive even numbers:
66+68+70+72+74+76+78+80+82+84+86+88+90+92
+94+96+98+100+102+104+106+108+110 = 2024.

2024 Magic Sums

2-0+2-4 = 0, so 2024 is divisible by 11. That means it is the magic sum of an 11 × 11 Magic Square. Here is one way that Magic Square can be completed. I followed the directions given in this post. You can see the 11 consecutive numbers listed above along the lower left to upper right diagonal.

If you would like to try completing the magic square yourself, here’s an Excel template that will automatically add the sums while you enter the numbers: 1766-1772 and 2024 Magic Squares

2024 is divisible by 8 but not by 16, so it is the Magic Sum of a 16 × 16 Magic Square. Here are two examples:

For this first one, I wrote the numbers from -1 to 14 across the top of the puzzle and continued in like manner. After the numbers were in place, I started flipping diagonals. How many diagonals did I flip?  34: The sixteen green diagonals, the sixteen blue diagonals, the pink diagonal, and the brown diagonal.

Here is how it looked when I finished:

For this second one, I wrote the numbers from -1 to 14 in the first 4 × 4 square, the numbers from 15 to 30 in the second 4 × 4 square, etc. Then I began flipping diagonals.

The Excel sheet, 1766-1772 and 2024 Magic Squares, also includes a template that will allow you to complete just a 4 × 4 magic square in the lower right corner for the numbers from -1 to 14. The whole 16 × 16 magic square will populate if you just complete that 4 × 4 magic square! But there’s a template if you want to do the whole thing from scratch as well.

More About the Number 2024:

Because 2024 is the sum of the 16 numbers from 119 to 134,
and 16 is even, it follows that
134²-133²+132²-131²+130²-129²+128²-127²+126²
-125²+124²-123²+122²-121²+120²-119²=2024.

Here is a way to make 2024 using only the digits 2, 0, 2, and 4:

(2+0+2+4)×
(2+0+2+4-(2+0)/2+4)×
(((2+0+2+4)×(-(2+0)/2+4))-(2+0+2)/4) = 2024.

The tan²(88.7266556386°) ≈ 2024, so it is the solution to this next problem:

 

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).

1770 This Christmas, Don’t Let the Taxman Get Most of Your Cash!

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Today’s Puzzle:

1770 = 30 · 59.
1770 = (60 · 59)/2.

That means 1770 is a triangular number. If we have 59 envelopes numbered 1 to 59, and each envelope contained the amount of money on the outside of the envelope, we would have $1770 in cash at stake. In this game, the TAXMAN wants to take as much money as he can get, but you control how much he can take: Can you allow him to get as little as possible? 

You can play Taxman easily with these printable Taxman “envelopes” and Taxman Scoring Calculator because each “envelope” lists all the factors of the envelope number. Your first selection should be the biggest prime number on the board because then the only envelope the taxman can get on that turn is the 1 envelope. The Taxman must be able to take at least one envelope on every turn. Try to make it so he can only get one or at most two envelopes on each turn. When it is no longer possible for you to take an envelope that allows the Taxman to take at least one envelope, too, the taxman gets ALL the rest of the envelopes. You win if you can keep more than half of your cash. Good luck!

Factors of 1770:

  • 1770 is a composite number.
  • Prime factorization: 1770 = 2 × 3 × 5 × 59.
  • 1770 has no exponents greater than 1 in its prime factorization, so √1770 cannot be simplified.
  • The exponents in the prime factorization are 1, 1, 1, and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 × 2 = 16. Therefore 1770 has exactly 16 factors.
  • The factors of 1770 are outlined with their factor pair partners in the graphic below.

More About the Number 1770:

As mentioned earlier 1770 is the 59th triangular number because 59(60)/2 = 1770.

1770 is also the 30th hexagonal number because 30(2·30-1) = 1770. (Every hexagonal number is also a triangular number.)

I’ve made images of hexagonal numbers before, but this time I wanted to make one using this hexagon template:

1770 is the hypotenuse of a Pythagorean triple:
1062-1416-1770, which is (3-4-5) times 354.

1770 is repdigit, UU, in base 58 because
30(58) + 30(1) = 30(58 + 1) = 30(59) = 1770.

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?