Number

1772 Is a Centered Heptagonal Number!

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Today’s Puzzle:

It’s early in 2024, so here’s a Factor Fits puzzle utilizing the factors of 20 and 24. Give it a try! There is only one solution.

Factors of 1772:

This is my 1772nd post. What are the factors of 1772?

  • 1772 is a composite number.
  • Prime factorization: 1772 = 2 × 2 × 443, which can be written 1772 = 2² × 443.
  • 1772 has at least one exponent greater than 1 in its prime factorization so √1772 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1772 = (√4)(√443) = 2√443.
  • 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 1772 has exactly 6 factors.
  • The factors of 1772 are outlined with their factor pair partners in the graphic below.

1772 is a Centered Heptagonal Number:

1772 is one more than 7 times the 22nd triangular number. For all previous centered heptagonal numbers about which I’ve written, I only mentioned their inclusion in this set of numbers. This time, I was determined to produce a graphic of the number. I used Desmos and Excel to determine all 1772 points in the graphic. It was a little time-consuming, but I got it done!

The points of the first heptagon were (1, 0), (cos2π/7, sin2π/7), (cos4π/7, sin4π/7), (cos6π/7, sin6π/7), (cos8π/7, sin8π/7), (cos10π/7, sin10π/7), (cos12π/7, sin12π/7). Here is an example of what was involved in completing one side of the other heptagons: Suppose I wanted to find five points on the line connecting (a,c) and (b,d). The five points would be
((4a+0b)/4, (4c+0d)/4), or simply (a, c),
((3a+1b)/4, (3c+1d)/4),
((2a+2b)/4, (2c+2d)/4), or simply ((a+b)/2, (c+d)/2), the midpoint,
((1a+3b)/4, (1c+3d)/4),
((0a+4b)/4, (0c+4d)/4), or simply (b, d).

I used Excel to calculate those numbers and then copied and pasted them into Desmos which graphed them beautifully. Each round took me less than ten minutes to complete. Here is the finished product:

More About the Number 1772:

1772 is the difference of two squares:
444² – 442² = 1772.

1772 is palindrome 24042 in base 5. Why?
Because 2(5⁴)+4(5³)+0(5²)+4(5¹)+2(5º) = 1772.

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.

1765 On This Memorial Day

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Today’s Puzzle:

This weekend I laid a bouquet of red and white flowers on my husband’s grave and decided to make a red rose Memorial Day puzzle for the blog as well. It is a mystery-level puzzle.

Write the number from 1 to 12 in both the first column and the top row so that those numbers are the factors of the given clues. There is only one solution.

Factors of 1765:

  • 1765 is a composite number.
  • Prime factorization: 1765 = 5 × 353.
  • 1765 has no exponents greater than 1 in its prime factorization, so √1765 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 1765 has exactly 4 factors.
  • The factors of 1765 are outlined with their factor pair partners in the graphic below.

More About the Number 1765:

1765 is the sum of two squares in two different ways:
42² + 1² = 1765, and
33² + 26² = 1765.

1765 is the hypotenuse of FOUR Pythagorean triples:
84 1763 1765, calculated from 2(42)(1), 42² – 1², 42² + 1²,
413 1716 1765, calculated from 33² – 26², 2(33)(26), 33² + 26²,
1059-1412-1765, which is (3-4-5) times 353, and
1125-1360-1765, which is 5 times (225-272-353).

1765 is a digitally powerful number:
1⁴ + 7³ + 6⁴ + 5³ = 1765.

1765 is a palindrome in a couple of different bases:
It’s A5A base 13 because 10(13²) + 5(13) + 10(1) = 1765, and
it’s 1D1 base 36 because 1(36²) + 13(36) + 1(1) = 1765.

1763 Daffodil Puzzle

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Today’s Puzzle:

Spring has sprung and perhaps flowers are blooming in your area. I think my favorite flowers are daffodils. I love the way they are shaped and their vibrant colors.

This daffodil puzzle is a great way to welcome spring. It may be a little bit tricky, but I think if you carefully use logic you will succeed! Just write each of the numbers 1 to 12 in the first column and again in the top row so that those numbers are the factors of the given clues. As always there is only one solution.

Here’s the same puzzle if you’d like to print it using less ink:

Factors of 1763:

  • 1763 is a composite number and the product of twin primes.
  • Prime factorization: 1763 = 41 × 43.
  • 1763 has no exponents greater than 1 in its prime factorization, so √1763 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 1763 has exactly 4 factors.
  • The factors of 1763 are outlined with their factor pair partners in the graphic below.

More About the Number 1763:

1763 is the difference of two squares in two different ways:
882² – 881² = 1763, and
42² – 1² = 1763. (That means the next number will be a perfect square!)

1763 is the hypotenuse of a Pythagorean triple:
387-1720-1763, which is (9-40-41) times 43.

1763 is palindrome 3E3 in base 22 because
3(22²) + 14(22) + 3(1) = 1763.

Lastly and most significantly: 15, 35, 143, 323, 899, and 1763 begin the list of numbers that are the product of twin primes. 1763 is just the sixth number on that list! If we include the products of two consecutive primes whether they are twin primes or not, the list is still fairly small. How rarely does that happen?

When it was 2021, did you realize how significant that year was?

1762 Happy Saint Patrick’s Day!

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Today’s ♣ Puzzle:

Here’s a much easier puzzle than yesterday’s for you to enjoy on this Saint Patrick’s Day. The diagonal lines on the corner boxes are only to help define the leaves of the shamrock.

Factors of 1762:

  • 1762 is a composite number.
  • Prime factorization: 1762 = 2 × 881.
  • 1762 has no exponents greater than 1 in its prime factorization, so √1762 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 1762 has exactly 4 factors.
  • The factors of 1762 are outlined with their factor pair partners in the graphic below.

More About the Number 1762:

1762 is the sum of two squares:
41² + 9² = 1762.

1762 is the hypotenuse of a Pythagorean triple:
738-1600-1762 calculated from 2(41)(9), 41² – 9², 41² + 9².
It is also 2 times (369-800-881).

1762 is palindrome 7C7 in base 15
because 7(15²) + 12(15) + 7(1) = 1762.

1761 Irish Harp

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Today’s Puzzle:

This mystery-level puzzle was meant to look a little like an Irish harp. Using logic write the numbers 1 to 12 in the first column and again in the top row so that those numbers and the given clues make a multiplication table. There is only one solution.

Notice that the clues 16 and 24 appear THREE times in the puzzle. In each case, you will need to determine if the common factor is 2, 4, or 8. You will have to get the common factor for each one in the right place or it will cause trouble for another clue. Consider what problems each of the following scenarios bring to other clues. For example, 48 must be either 4 × 12 or 6 × 8, but both possibilities are impossible in at least one of these scenarios:

Once you determine the only scenario that doesn’t present a problem for any other clue, you will be able to begin the puzzle.

Here’s the same puzzle without any added color:

Factors of 1761:

1 + 7 + 6 + 1 = 15, a number divisible by 3, so 1761 is divisible by 3. Since 6 is divisible by 3, we didn’t have to include it in our sum: 1 + 7 + 1 = 9, so 1761 is divisible by 3.

  • 1761 is a composite number.
  • Prime factorization: 1761 = 3 × 587.
  • 1761 has no exponents greater than 1 in its prime factorization, so √1761 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 1761 has exactly 4 factors.
  • The factors of 1761 are outlined with their factor pair partners in the graphic below.

More About the Number 1761:

1761 is the difference of two squares in two different ways:
881² – 880² = 1761, and
295² – 292² = 1761.

1761 is palindrome 1N1 in base 32
because 1(32²) + 23(32) + 1(1) = 1761.

1760 Pots of Gold and Rainbows

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Today’s Puzzle:

Write the numbers 1 to 12 in the first column and again in the top row so that those numbers are the factors that make the given clues. It’s a level 6, so it won’t be easy. Finding a leprechaun’s pot of gold isn’t easy either. Still, if you can solve this puzzle, then you will have found some real golden nuggets of knowledge.

They say there’s a pot of gold at the end of the rainbow. Where’s the rainbow?

Factors of 1760:

Puzzle number, 1760, has many factors. It makes a very big factor rainbow!

  • 1760 is a composite number.
  • Prime factorization: 1760 = 2 × 2 × 2 × 2 × 2 × 5 × 11, which can be written 1760 = 2⁵ × 5 × 11.
  • 1760 has at least one exponent greater than 1 in its prime factorization so √1760 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1760 = (√16)(√110) = 4√110.
  • The exponents in the prime factorization are 5, 1, and 1. Adding one to each exponent and multiplying we get (5 + 1)(1 + 1)(1 + 1) = 6 × 2 × 2 = 24. Therefore 1760 has exactly 24 factors.
  • The factors of 1760 are outlined with their factor pair partners in the graphic below.

More About the Number 1760:

1760 is the hypotenuse of a Pythagorean triple:
1056-1408-1760 which is (3-4-5) times 352.

1760 is the difference of two squares in eight different ways:
441² – 439² = 1760,
222² – 218² = 1760,
114² – 106² = 1760,
93² – 83² = 1760,
63² – 47² = 1760,
54² – 34² = 1760,
51² – 29² = 1760,  and
42² – 2² = 1760. (That means we are only four numbers away from the next perfect square!)

1760 is palindrome 2102012 in base 3
because 2(3⁶)+1(3⁵)+ 0(3⁴)+2(3³)+0(3²)+1(3¹)+2(3º) = 1760.

1759 Pie Over Two

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Today’s Puzzle:

Today in the United States many students will celebrate pi day by eating pie. Today’s puzzle looks a little like a pie that has been cut in half, so I’m calling it pie over two, abbreviated as “π/2”.

Write the numbers 1 to 12 in the first column and again in the top row so that those numbers and the given clues make a multiplication table. Be sure to use logic every step of the way.

Factors of 1759:

  • 1759 is a prime number.
  • Prime factorization: 1759 is prime.
  • 1759 has no exponents greater than 1 in its prime factorization, so √1759 cannot be simplified.
  • The exponent in the prime factorization is 1. Adding one to that exponent we get (1 + 1) = 2. Therefore 1759 has exactly 2 factors.
  • The factors of 1759 are outlined with their factor pair partners in the graphic below.

How do we know that 1759 is a prime number? If 1759 were not a prime number, then it would be divisible by at least one prime number less than or equal to √1759. Since 1759 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, or 41, we know that 1759 is a prime number.

More About the Number 1759:

Like every other odd number, 1759 is the difference of two squares:
880² – 879² = 1759.

OEIS.org informs us that 1759 is only the 17th Eisenstein-Mersenne prime number.

1758 Two-Shillelagh O’Sullivan

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Today’s Puzzle:

When I was looking for the song about the shillelagh for my previous post, I found another one called Two-Shillelagh O’Sullivan also by Bing Crosby. It wasn’t a song from my childhood, but it inspired me to make a puzzle with two shillelaghs anyway. In the song, O’Sullivan wears these walking sticks in a holster and can draw them quicker than anyone can draw a gun. He was impossible to beat.

This two-shillelagh puzzle is also a bit difficult to beat. You’re not going to let that stop you from trying, are you? Just use logic and your knowledge of the multiplication table.

Write the numbers from 1 to 12 in the first column and again in the top row so that those numbers are the factors of the given clues.

Here’s the same puzzle in black and white:

Factors of 1758:

  • 1758 is a composite number.
  • Prime factorization: 1758 = 2 × 3 × 293.
  • 1758 has no exponents greater than 1 in its prime factorization, so √1758 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 1758 has exactly 8 factors.
  • The factors of 1758 are outlined with their factor pair partners in the graphic below.

More About the Number 1758:

1758 is the hypotenuse of a Pythagorean triple:
408-1710-1758, which is 6 times (68-285-293).

1758 is palindrome 8E8 in base 14
because 8(14²) + 13(14) + 8(1) = 1758.