1652 Start at the Top and Work Your Way Down to the Bottom

Today’s Puzzle:

This is a level 3 puzzle so the clues are given in a logical order starting from the top of the puzzle. Begin by writing the factors of 20 and 32 in the appropriate cells. Then write the rest of the numbers so that both the first column and the top row have all the numbers from 1 to 10, and the written numbers are the factors of the given clues.

Factors of 1652:

  • 1652 is a composite number.
  • Prime factorization: 1652 = 2 × 2 × 7 × 59, which can be written 1652 = 2² × 7 × 59.
  • 1652 has at least one exponent greater than 1 in its prime factorization so √1652 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1652 = (√4)(√413) = 2√413.
  • 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 1652 has exactly 12 factors.
  • The factors of 1652 are outlined with their factor pair partners in the graphic below.

More About the Number 1652:

1652 is the difference of two squares two different ways:
414² – 412² = 1652 and
66² – 52² = 1652.

1651 Multiplication Fun

Today’s Puzzle:

Look how much fun these kids are having doing multiplication!

A game like that can help kids get ready to solve a fun puzzle based on the multiplication table.

Write each number from 1 to 10 in both the first column and the top row so that those numbers and the given clues become a multiplication table.

Factors of 1651:

  • 1651 is a composite number.
  • Prime factorization: 1651 = 13 × 127.
  • 1651 has no exponents greater than 1 in its prime factorization, so √1651 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 1651 has exactly 4 factors.
  • The factors of 1651 are outlined with their factor pair partners in the graphic below.

More About the Number 1651:

1651 is the hypotenuse of a Pythagorean triple:
635-1524-1651, which is (5-12-13) times 127.

1651 is the 26th heptagonal number because
5(26²)/2 – 3(26)/2 = 1651.

1651 is a nice-looking palindrome in base 2:
1651₁₀ = 11001110011₂.
That just means that
2¹⁰ + 2⁹ + 2⁶ + 2⁵ + 2⁴+ 2¹+ 2⁰ = 1024 + 512 + 64 + 32 + 16 + 2 + 1 = 1651.

 

 

1650 Wrinkles in the Multiplication Table

Today’s Puzzle:

Are you familiar with the book A Wrinkle in Time? Kat of The Lily Cafe’s blog loves books and recently compared Meg in that book to her six-year-old son. She wrote a post titled Am I Raising a Meg? Her six-year-old LOVES math and is very much interested in multiplication and division. When Mom thought he was playing a game on her phone, he was actually playing with the calculator app! I felt so happy inside as I read that!

I wonder if they have discovered the storybooks in the Math Book Magic blog. Such books could combine Mom’s love for reading with her son’s love of math.

Someday her son might like to solve a “wrinkled” multiplication table puzzle like this one that has only nine clues.

Write all the numbers 1 to 10 in both the first column and the top row so that those numbers and the given clues become a multiplication table.

Factor Cake for 1650:

This is my 1650th post.
1650 is divisible by 2 and by 5 because it ends with a 0.
1650 is divisible by 3 because 1 + 6 + 5 + 0 = 12, a number divisible by 3.
1650 is divisible by 11 because 1 – 6 + 5 – 0 = 0, a number divisible by 11.

I think we can make a lovely factor cake for 1650:

Factors of 1650:

  • 1650 is a composite number.
  • Prime factorization: 1650 = 2 × 3 × 5 × 5 × 11, which can be written 1650 = 2 × 3 × 5² × 11.
  • 1650 has at least one exponent greater than 1 in its prime factorization so √1650 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1650 = (√25)(√66) = 5√66.
  • The exponents in the prime factorization are 1, 1, 2, and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1)(2 + 1)(1 + 1) = 2 × 2 × 3 × 2 = 24. Therefore 1650 has exactly 24 factors.
  • The factors of 1650 are outlined with their factor pair partners in the graphic below.

More About the Number 1650:

1650 is the hypotenuse of TWO Pythagorean triples:
462-1584-1650, which is (7-24-25) times 66, and
990-1320-1650, which is (3-4-5) times 330.

1649 Tweaking a Puzzle Posed by Sunil Singh @Mathgarden

Today’s Puzzle:

A couple of days ago on Twitter, I saw an interesting puzzle posed by Sunil Singh @Mathgarden.

After I found one of several of its solutions, I wondered if I could add a bridge that would use all nine numbers from 1 to 9 in the solution, so I tweaked it. I decided to move the puzzle to the ocean when I added that extra bridge.

I was able to solve this problem using logic and addition facts, rather than algebra. Try solving it yourself. If you want to see any of the steps I used to solve the puzzle, scroll down to the end of the post.

Factors of 1649:

  • 1649 is a composite number.
  • Prime factorization: 1649 = 17 × 97.
  • 1649 has no exponents greater than 1 in its prime factorization, so √1649 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 1649 has exactly 4 factors.
  • The factors of 1649 are outlined with their factor pair partners in the graphic below.

More About the Number 1649:

1649 is the sum of two squares in TWO different ways:
32² + 25² = 1649, and
40² + 7² = 1649.

1649 is the hypotenuse of FOUR Pythagorean triples:
399-1600-1649, calculated from 32² – 25², 2(32)(25), 32² + 25²,
560-1551-1649, calculated from 2(40)(7), 40² – 7², 40² + 7²,
776 1455 1649, which is (8-15-17) times 97, and
1105 1224 1649, which is 17 times (65-72-97).

Some Logical Steps to Solve Today’s Puzzle:

I found four different ways to write the solution but they are all rotations or reflections of each other. Here are the steps to one of those four ways:

1st Step: The biggest number that can be used is 9. Since every island must be included in at least two sums, neither 8 nor 9 can be an addend; they both must be sums. 7 must be an addend in their sums because adding any number to 7 will yield 8, 9, or some larger forbidden number. Thus 8 and 9 are bridges that connect to island 7.

I chose to write 7, 8, and 9 in the top right section, but 8 and 9 could change places with each other. I could have also chosen to write those numbers in the bottom left area.

2nd Step: 1 + 7 = 8, and 7 + 2 = 9.

3rd Step: 1 + 2 = 3.

4th Step: The last island must be the smallest remaining number (4) because the smallest remaining number can’t be the sum of a bigger number and the number on either adjacent island.

Final Step: 1 + 4 = 5, and 4 + 2 = 6.

Did you enjoy this puzzle? How did my steps compare to the steps that you took?

Please, check the comments for another solution.

1648 A Pythagorean Triple Logic Puzzle with a Triangular Card Deck

Today’s Puzzle:

It’s been a few years since I’ve made one of these Pythagorean triple logic puzzles. The triangles in it are shaped a little different than in years past because I also wanted to make a deck of Pythagorean triple cards that are shaped like equilateral triangles or at least as close as I can get to equal sides. This puzzle won’t be easy, but do give it a try!

PUZZLE DIRECTIONS: This puzzle is NOT drawn to scale. Although all of the angles may look like 60-degree angles, none of them are. The marked angles are 90 degrees. Lines that look parallel are NOT parallel. Although side lengths look equal, they are NOT equal. Most rules of geometry do not apply here: in fact, non-adjacent triangles in the drawing might actually overlap.

No geometry is needed to solve this puzzle. All that is needed is logic and the table of Pythagorean triples under the puzzle. The puzzle only uses triples in which each leg and each hypotenuse is less than 100 units long. The puzzle has only one solution.

Sorted Triples

Print the puzzles or type the solution in this excel file: 12 Factors 1639-1648

Triangular Card Deck of Pythagorean Triples:

You can make this deck of 50 playing cards to help you solve the puzzle or perhaps to play a domino-type game. Print each group of 25 cards on a separate sheet of paper. Cut the cards out along the solid lines and fold the cards on the dotted lines. Use a glue stick to keep the front of each card attached to its back. Laminate the cards, if desired. The right angles are the only angles marked on the cards, and the hypotenuses are all along the folded edges. Note: Some sides will not match with any other side in the deck, and the 57-76-95 triangle does not match sides with ANY triangle with a hypotenuse less than 100.

Factors of the Number 1648:

  • 1648 is a composite number.
  • Prime factorization: 1648 = 2 × 2 × 2 × 2 × 103, which can be written 1648 = 2⁴ × 103.
  • 1648 has at least one exponent greater than 1 in its prime factorization so √1648 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1648 = (√16)(√103) = 4√103.
  • The exponents in the prime factorization are 4 and 1. Adding one to each exponent and multiplying we get (4 + 1)(1 + 1) = 5 × 2 = 10. Therefore 1648 has exactly 10 factors.
  • The factors of 1648 are outlined with their factor pair partners in the graphic below.

More About the Number 1648:

1648 is not the hypotenuse of any Pythagorean triples because none of its prime factors leave a remainder of 1 when divided by 4.

1648 is the difference of two squares three different ways:
413² – 411² = 1648,
208² – 204² = 1648, and
107² – 99² = 1648.

That means 1648 is a leg in THREE Pythagorean triples calculated from
413² – 411², 2(413)(411), 413² + 411²;
208² – 204², 2(208)(204), 208² + 204²; and
107² – 99², 2(107)(99), 107² + 99².

1648 can be expressed as 2(824)(1), 2(412)(2), 2(206)(4), as well as 2(103)(8).

That means 1648 is a leg in FOUR Pythagorean triples calculated from
2(824)(1), 824² – 1², 824² + 1²;
2(412)(2), 412² – 2², 412² + 2²;
2(206)(4), 206² – 4², 206² + 4²; and
2(103)(8), 103² – 8², 103² + 8².

Sometimes those formulas produce duplicate triples, but not this time. 1648 is in SEVEN Pythagorean triples!
1648-339486-339490,
1648-84864-84880,
1648-21186-21250,
1648-678975-678977,
1648-169740-169748,
1648-42420-42452,
1648-10545-10673.

Aren’t you glad that today’s puzzle was limited to triples with hypotenuses less than 100? I certainly wouldn’t want any of those triples with 1648 as a side to be part of any puzzle! But 2-digit sides, 16 and 48, helped to make today’s puzzle, and I do hope you were able to solve it today!

Yes, YOU Can Host a Playful Math Education Blog Carnival

I had so much fun hosting the 146th Playful Math Education Blog Carnival. Kelly Darke of Math Book Magic will host the 148th Carnival. Probably neither of us should host the 147th Carnival, but YOU most certainly can! By YOU I mean anyone who has ever blogged even just a little bit about math. For example, if you normally blog about art, you could create a carnival that mostly focuses on mathematical art. The same could be said for photography, games, puzzles, storybooks, and so forth.

If now isn’t a good time for you to take on an extra project, remember there are plenty of other months open for you to volunteer!

But how do you host the Playful Math Education Blog Carnival, you ask? First of all, let Denise Gaskins know you would like to host the carnival.

You can also contact her through Twitter:

After you get assigned a month and a carnival number, you should pick a day in the last full week of the month as your goal to publish your carnival.

You might be interested in knowing how I approach creating a carnival:

Number Facts or a Puzzle:

Traditionally you start with some facts or a puzzle about the current carnival number. You can find several facts about your number at Pat Ballew’s Math Day of the Year Facts, Wikipedia, or OEIS.org. Also, check The Carnival of Mathematics which is about 48 numbers ahead of the Playful Math Education Carnival. What interesting facts were written about your number around four years ago in that carnival? You don’t have to be fancy; you can simply state a fact or two about your carnival number.

I, on the other hand, am obsessive. If I were hosting the 147th Carnival I would find as many facts about the number 147 as I could. I would think about all those facts and try to come up with a way to marry my number with something about a carnival, a fair, or even a circus. After a couple of weeks of imagining, I would finally be able to tell you about the great contortionist, Hexahex. Perhaps you’ve heard of his mother, Polly Hex. Hexahex can contort himself into 82 different “free” positions. He wants to stretch himself a little bit and add 65 more “one-sided” positions for a total of 147 “one-sided” positions in his repertoire. He is allowed to count positions that are reflections of the first 82 positions, but only if they aren’t exactly the same or merely a rotation of any of those first 82. Below is a graphic showing those first 82 positions as well as their reflections. Put an X above the 17 positions in the bottom three rows that don’t qualify as different, then count up the rest. You will then see that Hexahex can indeed contort himself 147 ways!

See, I told you I am obsessive! If you host the 147th carnival, you can use my graphic and story about Hexahex if you like. If you don’t want to use it, that’s okay, too!

As Denise Gaskins advised,

You decide how much effort you want to put in. Writing the carnival can take a couple of hours for a simple post, or you can spend several days searching out and polishing playful math gems to share.

I try to start writing a draft of my blog carnival post long before my deadline. I collect pictures (good advice on finding pictures here) and quotations whenever I find something I like, and enter them into my post ahead of time. If I have the framework in place, then all I have to add at the last minute are the blog post links, and the job doesn’t seem overwhelming.

Make sure you have the right to use any image you post. Either create a graphic yourself or find something marked “Creative Commons” — and then follow the CC rules and give credit to the artist/photographer.

I typically use graphics I’ve made or embed tweets from Twitter that just seem to have the perfect picture or quote.

Finding Blog Posts for Your Carnival Through Your Blog’s Reader:

Second, you look for blog posts. I found some blog posts because I subscribe to them, but you can also find blog posts by searching your reader. You may think blogging is dead, but it most certainly isn’t. I blog on WordPress, and its reader is easy to search. The search terms I used included math art, math poetry, math games, math puzzles, math geometry, and math algebra. Here are blog posts I found recently, most of which were written after the last carnival was published. Others were written before my carnival, but somehow I missed finding them before. If you hosted the 147th carnival, you will want to organize the posts into different categories or age groups and write a brief introduction to each post, but you could include as many or as few of these posts as your heart desires as well as other posts that you find. Here’s a bonus: if you also blog on WordPress, as soon as you hit the submit button, then WordPress will let the authors know that their post was included in your carnival! I have not organized these blog posts, but click on any of them that look interesting to you and consider including them in your carnival. If they don’t look interesting, a good introduction written by you might make them appeal to more people.

Finding Blog Posts on Twitter:

Twitter has SO many wonderful, playful ideas about mathematics. Most of them do not come from blog posts, but some of them do. Often when I see a tweet that refers to a math blog post or something else I like, I hit the like button. You can check my Twitter profile to see what appeals to me. Twitter also has a search feature. I’ve searched for individuals that I know who blog. I’ve also used words like “math blog” in my Twitter search to find blog posts I haven’t seen before. Be aware that you may find posts that are old or have no date on them, but plenty of recent blog posts are just waiting for you to find! Also, Denise Gaskins will retweet some blog posts that she’s found. Here are the blog posts I found on Twitter AFTER my carnival was published. Again, if you were hosting the carnival, the posts to include in your carnival would be up to you. You would organize them into different categories or age groups and write a brief introduction for each post.

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That blog post doesn’t appear to be recent, but it did lead me to this one published in May 2021: Accelerate Vs. Remediate.

Embedding tweets on your blog can make the post seem VERY long. I would select a few of my favorite tweets with pictures to embed in the carnival and just use the blog links to take my readers directly to most of the posts.

You can also post a link to your carnival on Twitter with a thank you acknowledging the Twitter handles of people whose blog posts you used. That isn’t a required step, but it will help to get the word out to more people to visit your carnival.

Some Final Steps:

After you’ve organized all the blog posts into different categories or age groups and written briefly about them, stop looking for more blog posts, because there will always be more, and if you don’t stop looking, you will never be finished! It is a good idea to make sure the links you’ve included really do take your readers where you think you are sending them. I admit that I’ve messed up on that detail before.

To finish up, you will want to include a link to the previous playful math carnival and a link to the website of the next carnival, if known. You can find that information here. You will want to include an invitation for others to host future carnivals. It is also courteous to direct your readers to the current edition of the Carnival of Mathematics. Lastly, proofread and publish! Good luck and have fun!

 

1647 A Fun Mystery

Today’s Puzzle:

I’ve never made a puzzle quite like this one before, and even though the logic needed is a little tricky, I found it quite enjoyable. Share it with a friend and see if you don’t have some great discussions such as if the common factor of 12 and 36 is 3, 4, 6, or 12.

Factors of 1647:

1 + 6 + 4 + 7 = 18, a multiple of 3 and 9, so 1647 is divisible by both 3 and 9.

  • 1647 is a composite number.
  • Prime factorization: 1647 = 3 × 3 × 3 × 61, which can be written 1647 = 3³ × 61.
  • 1647 has at least one exponent greater than 1 in its prime factorization so √1647 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1647 = (√9)(√183) = 3√183.
  • 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 1647 has exactly 8 factors.
  • The factors of 1647 are outlined with their factor pair partners in the graphic below.

More About the Number 1647:

1647 is the hypotenuse of a Pythagorean triple:
297-1620-1647, which is 27 times (11-60-61).

1647 is a repdigit in base 13:
1647₁₀ = 999₁₃ because
9(13² + 13¹ + 13º) =
9(169 + 13 + 1) =
9(183) = 1647.

1646 Mystery Level

Today’s Puzzle:

It’s a mystery if this puzzle is easy, difficult, or half easy and half difficult. The only way for you to know is to use logic to try solving it yourself!

Factors of 1646:

  • 1646 is a composite number.
  • Prime factorization: 1646 = 2 × 823.
  • 1646 has no exponents greater than 1 in its prime factorization, so √1646 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 1646 has exactly 4 factors.
  • The factors of 1646 are outlined with their factor pair partners in the graphic below.

More About the Number 1646:

823 is part of the prime decade: 821, 823, 827, 829.
You can be sure that each of those primes doubled: 1642, 1646, 1654, and 1658 will have exactly four factors.

1646 is in only one Pythagorean triple:
1646-677328-677330, calculated from 2(823)(1), 823² – 1², 823² + 1².

1645 and Level 6

Today’s Puzzle:

The number 36 appears as a clue in this puzzle three times. None of those 36’s will be 3 × 12 because 21 and 33 must use both 3’s. That means two of the 36’s will be 4 × 9, and one of them will be 6 × 6. Can both of the 36’s associated with the 24 be 4 × 9? Answering that question will help you find the logic needed to know which common factor to use for 72 and 36.

Factors of 1645:

  • 1645 is a composite number.
  • Prime factorization: 1645 = 5 × 7 × 47.
  • 1645 has no exponents greater than 1 in its prime factorization, so √1645 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 1645 has exactly 8 factors.
  • The factors of 1645 are outlined with their factor pair partners in the graphic below.

 

More About the Number 1645:

1645 is the hypotenuse of a Pythagorean triple:
987-1316-1645, which is (3-4-5) times 329.

1644 Level 5 Puzzles Are Not So Easy to Solve

Today’s Puzzle:

This puzzle isn’t so easy to solve. For example, the common factor of clues 48 and 24 might be 4, 6, 8, or 12. Which one should you use? Logic will help answer that question. Give this puzzle a try?

Factors of 1644:

1644 has 12 factors and is divisible by 12.

  • 1644 is a composite number.
  • Prime factorization: 1644 = 2 × 2 × 3 × 137, which can be written 1644 = 2² × 3 × 137.
  • 1644 has at least one exponent greater than 1 in its prime factorization so √1644 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1644 = (√4)(√411) = 2√411.
  • 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 1644 has exactly 12 factors.
  • The factors of 1644 are outlined with their factor pair partners in the graphic below.

More about the Number 1644:

1644 is the hypotenuse of a Pythagorean triple:
1056-1260-1644 which is 12 times (88-105-137).