Easter Basket Challenge

Occasionally,  we hear that the number of Easter eggs that are found is one or two less than the number of eggs that were hidden. Still most of the time, all the eggs and candies do get found. You really have no trouble finding all those goodies, and the Easter Egg Hunt seems like it is over in seconds.  You can find Easter Eggs but can you find factors? Here’s an Easter Basket Find the Factors 1 – 10 Challenge Puzzle for you. I guarantee it won’t be done in seconds. Can you find all the factors? I dare you to try!


Let’s Get Ready for the Playful Math Carnival!

Too many people think that mathematics is a house of horrors, but there are plenty of bloggers out there, who know that done right, math is actually ALL fun and games. It is like a carnival! Every month, you can play at the Playful Math Education Blog Carnival, and it really is play! What does a playful math carnival look like? Go on over to see how Math Mama Writes… and puts on a fabulous March carnival!

I will be hosting this monthly carnival the last week of April! Why do I get to host it? I sent a message on twitter to Denise Gaskins who coordinates the carnival, and I requested the privilege. If you would like to host it in the future, let her know. She is always looking for blogs to host, and she will be very happy to hear from you.

In the meantime, you can help me with my carnival. If you blog about mathematics in a playful way that could benefit children who are somewhere between preschool to high school age, I would love to include your post in my carnival. The carnival is a FREE way to promote your post, so if you would like more traffic to your blog, submit a post using the link from Denise Gaskins’ website by Friday, April 19. Then before the end of the month, you will be able to enjoy the carnival even more because of your participation!

Yahtzee How-Many-Rolls Variation

Today on the spot I made up a quick variation of Yahtzee, and one of my students played it with me.

The object of the game was to get all five dice to show the same number of dots at the same time, but instead of only being allowed to have up to three rolls, we took as many rolls as need. To take a turn, one of us would roll the dice then look to see if any of the dice were the same. Any die that didn’t match would be included in an additional roll until it did match. We counted each roll we took and got one point for each roll. The lowest score would determine the winner. The student and I played four rounds. He was elated because he won EVERY round so, of course, he was the overall winner, too.

Usually, when we play a game together the scores are much closer. Sometimes I win, sometimes he wins. Today I couldn’t believe my bad luck! Sometimes none of the dice matched after my first roll. And what about my student’s very good luck getting five of a kind in just one roll? I’m sure some good probability discussions could result from this game.

Our data might suggest that 9 rolls is the most that a person could get, but I rolled the dice for the picture included in this post, and it took me 19 rolls to get that Yahtzee! And I actually had four 4’s after just 6 rolls before I took those last 13 rolls.

You never know for sure what will happen when it comes to games of chance. If you study probability, you can have a good idea about what is most likely to happen, but you cannot guarantee it will happen. If we had taken the time to play more rounds, maybe my student would have needed 10 or more rolls to get at least one of his Yahtzees, and the game would have been more competitive. (At least, that was what I was thinking before he rolled on rounds 3 and 4.)

I’d like to encourage you to try playing this game, too. I thought it was a lot of fun even though I lost miserably.


Today is a Good Day to Review Proof by Induction

0² = 0
1²  = 1

Does that pattern hold for all natural numbers? Could we claim that n²  = n?

Yes, we can, and I’ve written a proof to prove it! The proof uses a valuable concept in mathematics called induction. I remember being introduced to proofs by induction when I was in Junior High. Nowadays, if it is not part of Common Core, it wouldn’t be taught much anymore. Nevertheless, I will use it here to prove that n² = n.

Using a similar proof, we can also prove that n³ = n, n⁴ = n, n⁵ = n, n⁶ = n, and so forth!

Today is the perfect day to review how to use proof by induction so try your hand at proving at least one of those mathematical statements on your own. Use the same steps in my example: prove true for n=1, assume true for n = k, prove true for k + 1, write your conclusion. then have a very Happy April Fools’ Day, Everyone!

Today is also a very good day to review that (x + y)² = x² +2xy + y²  and NOT x²  + y², a very common error students make. Confession: I remember making that exact error in high school when I definitely should have known better. Using induction to prove something in mathematics is a valid technique, but if you use invalid equations like
(x + y)³ = x³ + y³, you will make invalid conclusions. Thus, today might also be a good day to review the binomial theorem and Pascal’s triangle. (Pascal’s triangle has numbers in its interior, not just 1’s going down the sides, after all.)

My post today was inspired by a post written by Sara Van Der Werf titled Why I’ve Started Teaching the FOIL Method Again. In her post, she not only plays a great April Fools’ joke on her readers, but she explains a tried and true way to multiply binomials and other polynomials.

I read her post exactly one year ago today, and since then, I have been waiting for April Fools’ Day to roll around again so that I could share this post with you. It is my hope that you will enjoy my little prank and learn a little mathematics from it as well.

Detail Left Out of the History Books

Today I was indexing some July 1944 death records from Budapest, Hungary and noticed that Boldizsár Klein and his wife, Regina Leichtmann, died only one day apart from each other. We don’t index causes of death, but I looked at their causes of death because their deaths were so close to each other. The same word was used for both causes of death. I wasn’t sure of all the letters in the word, but it started the same as a word I had seen before, öngyilkos, which literally means self-murder.

First I consulted my hardback Hungarian dictionary, but I didn’t find the word. Next, I looked at two online Hungarian genealogy dictionaries. Finally, I typed what the letters most looked like to me into Google Translate. After a few trials and errors with different letters of the alphabet with and without the prefix, ön, I found the word and their cause of death, önmérgezés, which means self-poisoning or intoxication.

Why did this happen to them?!!

From the record, I knew that both 74-year-old Boldizsár and 66-year-old Regina were Jewish. I googled and learned that the Nazis invaded its previous ally, Hungary, only a few months earlier on 19 March 1944 and mass evacuation of Jews to death camps began immediately. Since this couple lived in Budapest, the horrors of this occupation must have been felt most intensely. I cannot imagine what they went through, but trying to put ourselves in their shoes may help prevent history from repeating itself.

1366 Fractions Acting Improperly

In elementary school, we learned about improper fractions. Should we call them that? Is it even possible for any kind of number to be IMPROPER? They are simply fractions greater than one. I’ve recently heard the term fraction form used, and ever since I’ve made a point of saying that fractions greater than one are in fraction form.

On Twitter, I’ve found a few people who also don’t like using the word improper to describe any fraction.

This first tweet has a link explaining why it is improper to use the term improper fraction:

I hope that you will consider not labeling any fraction as improper, as well!

Now I’ll write a little bit about the number 1366:

  • 1366 is a composite number.
  • Prime factorization: 1366 = 2 × 683
  • The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1366 has exactly 4 factors.
  • Factors of 1366: 1, 2, 683, 1366
  • Factor pairs: 1366 = 1 × 1366 or 2 × 683
  • 1366 has no square factors that allow its square root to be simplified. √1366 ≈ 36.95944

1366 is also the sum of the twenty-six prime number from 5 to 107. Do you know what all those prime numbers are?

1365 Shamrock Mystery

Beautiful shamrocks with their three heart-shaped leaves are not difficult to find. Finding the factors in this shamrock-shaped puzzle might be a different story.  Sure, it might start off to be easy, but after a while, you might find it a wee bit more difficult, unless, of course, the luck of the Irish is with you!

Now I’ll share some information about the number 1365:

  • 1365 is a composite number.
  • Prime factorization: 1365 = 3 × 5 × 7 × 13
  • The exponents in the prime factorization are 1, 1, 1, and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 × 2 = 16. Therefore 1365 has exactly 16 factors.
  • Factors of 1365: 1, 3, 5, 7, 13, 15, 21, 35, 39, 65, 91, 105, 195, 273, 455, 1365
  • Factor pairs: 1365 = 1 × 1365, 3 × 455, 5 × 273, 7 × 195, 13 × 105, 15 × 91, 21 × 65, or 35 × 39
  • 1365 has no square factors that allow its square root to be simplified. √1365 ≈ 36.94591

1365 is the hypotenuse of FOUR Pythagorean triples:
336-1323-1365 which is 21 times (16-63-65)
525-1260-1365 which is (5-12-13) times 105
693-1176-1365 which is 21 times (33-56-65)
819-1092-1365 which is (3-4-5) times 273

1365 looks interesting in some other bases:
It’s 10101010101 in BASE 2,
111111 in BASE 4,
2525 in BASE 8, and
555 in BASE 16

I’m feeling pretty lucky that I noticed all those fabulous number facts! If you haven’t been so lucky finding the factors of the puzzle, the same puzzle but with more clues might help:

1350 Logic is at the Heart of This Puzzle

By simply changing two clues of that recently published puzzle that I rejected, I was able to create a love-ly puzzle that can be solved entirely by logic. Can you figure out where to put the numbers from 1 to 12 in each of the four outlined areas that divide the puzzle into four equal sections? If you can, my heart might just skip a beat!

Now I’ll tell you a few things about the number 1350:

  • 1350 is a composite number.
  • Prime factorization: 1350 = 2 × 3 × 3 × 3 × 5 × 5, which can be written 1350 = 2 × 3³ × 5²
  • The exponents in the prime factorization are 1, 3 and 2. Adding one to each and multiplying we get (1 + 1)(3 + 1)(2 + 1) = 2 × 4 × 3 = 24. Therefore 1350 has exactly 24 factors.
  • Factors of 1350: 1, 2, 3, 5, 6, 9, 10, 15, 18, 25, 27, 30, 45, 50, 54, 75, 90, 135, 150, 225, 270, 450, 675, 1350
  • Factor pairs: 1350 = 1 × 1350, 2 × 675, 3 × 450, 5 × 270, 6 × 225, 9 × 150, 10 × 135, 15 × 90, 18 × 75, 25 × 54, 27 × 50 or 30 × 45
  • Taking the factor pair with the largest square number factor, we get √1350 = (√225)(√6) = 15√6 ≈ 36.74235

1350 is the sum of consecutive prime numbers two ways:
It is the sum of the fourteen prime numbers from 67 to 131, and
673 + 677 = 1350

1350 is the hypotenuse of two Pythagorean triples:
810-1080-1350 which is (3-4-5) times 270
378-1296-1350 which is (7-24-25) times 54

1350 is also the 20th nonagonal number because 20(7 · 20 – 5)/2 = 1350

1349 A Rejected Puzzle

I was in the mood to make a Find the Factors Challenge Puzzle that used the numbers from 1 to 12 as the factors. I’ve never made such a large puzzle before, but after I made it, I rejected it. All the puzzles I make must meet certain standards: they must have a unique solution, and that solution must be obtainable by using logic. Although the “puzzle” below has a unique solution, and you can fill in a few of the cells using logic, you would have to use guess and check to finish it. Besides that, you wouldn’t be able to know if you guessed right until almost the entire puzzle was completed. Thus, it doesn’t meet my standards.

Even though the puzzle was rejected, there were still some things about it that I really liked. In my next post, I’ll publish a slightly different puzzle that uses some of the same necessary logic that I appreciated but doesn’t rely on guess and check at all. This is NOT the first time I have tweaked a puzzle that didn’t initially meet my standards to make it acceptable. I just thought I would share the process this time. If you try to solve it, you will be able to see the problem with the puzzle yourself.

Now I’ll share some information about the number 1349:

1349 is the sum of 13 consecutive primes, and it is also the sum of three consecutive primes:
73 + 79 + 83 + 89 + 97 + 101 + 103 + 107 + 109 + 113 + 127 + 131 + 137 = 1349
443 + 449 + 457 = 1349

1348 Coloring Paula Krieg’s Polar Rose

Paula Beardell Krieg recently wrote about using Desmos to create designs that can be colored by hand or by computer programs like Paint. I like using Paint so with her permission I took a design she made and colored it so I could present it here in this post. I chose colors that make me think of spring because, frankly, I’m ready for winter to be over!

Now I’ll write a little bit about the number 1348:

1348 is the sum of two squares:
32² + 18² = 1348

1348 is the hypotenuse of a Pythagorean triple:
700-1152-1348 which is 32² – 18², 2(32)(18), 32² + 18²

1348 is also the short leg in a primitive Pythagorean triple: