Posts tagged ‘puzzle’
This is what kept me up last night. Literally.
Form a row of ten squares, with each square randomly colored red, green, or yellow. Call this Row 1. Then place nine squares in Row 2 slightly offset above Row 1, and color the squares in Row 2 according to the following rules:
- If two side-by-side squares have the same color, the square between them in the row above has the same color.
- If two side-by-side squares have different colors, the square between them in the row above has the third color.
Continue in this manner with eight squares in Row 3, seven squares in Row 4, and so on, with just one square in Row 10.
Here’s an example of a pyramid constructed in this manner:
The 1’s, 2’s and 3’s in the diagram appear because this image was created with Microsoft Excel. Formulas were used to determine the number in each square, and conditional formatting was used to color the squares.
So far, this is just a test of how well you can follow rules, and it isn’t much fun. But here’s where it gets interesting.
How can you predict the color of the lone square in Row 10 after seeing only the arrangement of squares in Row 1 (and without constructing the rows in between)?
I found this problem last night on the Purdue Math Department’s Problem of the Week website. I lay awake in bed longer than I should have, but no solution came to me either while laying there awake or while I was sleeping. When I woke up, I shared it with my sons. I suspected they’d have fun coloring squares; I never suspected what actually happened.
I explained the problem to Alex and Eli, and I showed them an example of how to generate Row 9 from an arbitrary Row 10. They then pulled out their box of crayons and constructed rows 8, 7, 6, …, 1. Alex then looked at his pyramid for about 8 seconds and said, “Oh, I get it. You can find the color of the top square by _________.” (Spoiler omitted.)
“Is that your conjecture?” I asked.
“What’s a conjecture?” he replied.
“It’s a guess,” I told him. “It’s what you think the rule is.”
“No,” he said. “It’s not a guess. That’s the rule.”
He was pretty cocky for a seven-year-old.
So we tested his rule for another randomly-generated Row 10. It worked. So we tested it again, and it worked again. We tested it for six different hand-drawn pyramids… and it worked for every one of them.
That’s when I generated this Excel file. We used it to test Alex’s conjecture on 100+ other pyramids. It worked every time.
I still have no idea how he divined the rule so quickly.
For me, though, the cool part came when I was able to extend the puzzle with the following:
For what values of k can you predict the color of the lone square in Row k when there are k blocks in Row 1?
Trivially, if I gave you just two blocks in Row 1, you could most certainly predict the color of the square in Row 2.
But if I gave you, say, five blocks in Row 1, could you predict the color of the lone square in Row 5?
The final part of the Puzzle of the Week description from the Purdue website says exactly what you’d expect it to say:
Prove your answer.
I have a proof showing that Alex’s conjecture holds. Incidentally, that proof can be extended to prove the general result for the extension just posed.
Alex, however, has not yet generated a proof of his conjecture.
Then again, he’s only seven.
Saying that I like KenKen® would be like saying that Sigmund Freud liked cocaine. (Too soon?) ‘Twould be more proper to say that I am so thoroughly addicted to the puzzle that the length of my dog’s morning walks aren’t measured in miles or minutes but in number of 6 × 6 puzzles that I complete. (Most mornings, it’s two.) Roberto Clemente correctly predicted that he would die in a plane crash; Abraham de Moivre predicted that he would sleep to death (and the exact date on which it would occur… creepy); and I am absolutely certain that I’ll be hit by oncoming traffic as I step off the curb without looking, my nose pointed at a KenKen app on my phone and wondering, “How many five-element partitions of 13 could fill that 48× cell?”
I am forever indebted to Tetsuya Miyamoto for inventing KenKen, and I am deeply appreciative that Nextoy, LLC, brought KenKen to the United States. How else would I wile away the hours between sunrise and sunset?
I am also extremely grateful that the only thing Nextoy copyrighted was the name KenKen. This allows Tom Snyder to develop themed TomToms, and it allows the PGDevTeam to offer MathDoku Pro, which I believe to be the best Android app for playing KenKen puzzles.
The most recent release of MathDoku has improved numerical input as well as a timer. Consequently, my recent fascination is playing 4 × 4 puzzles to see how long it will take. A typical puzzle will take 20‑30 seconds; occasionally, I’ll complete a puzzle in 18‑19 seconds; and, every once in a while, I’ll hit 17 seconds… but not very often.
Today, however, was a banner day. I was in a good KenKen groove, and I was served one of the easiest 4 × 4 puzzles ever. Here’s the puzzle:
And here’s the result (spoiler):
The screenshot shows that I completed the puzzle in just 15 seconds. And it’s not even photoshopped.
This puzzle has several elements that make it easy to solve:
- The [11+] cell can only be filled with (4, 3, 4).
- The  in the first column dictates the order of the (1, 4) in the [4×] cell.
- The (1, 4) in the [4×] cell dictates the order of the (1, 2) in the [3+] cell.
After that, the rest of the puzzle falls easily into place, because each digit 1‑4 occurs exactly once in each row and column.
What’s the fastest you’ve ever solved a 4 × 4 KenKen puzzle? Post your time in the comments. Feel free to post your times for other size puzzles, too. (I’m currently working on a 6 × 6 puzzle that’s kicking my ass. Current time is 2:08:54 and counting.)
Here’s a math puzzle that is rather easy. Or is it?
You look in a mirror and see the reflection of a clock. In the reflection, the clock appears to show a time of 11:51. What is the real time?
Before I share the solution, how ’bout some clock jokes?
A hungry clock goes back four seconds.
I spent 35 minutes fixing a broken clock yesterday.
At least, I think it was 35 minutes…
What’s the difference between a man and a broken clock?
At least a broken clock is right twice a day.
It took me four hours to eat a dozen clocks.
It was very time consuming.
An alarm clock made of herbs will help you wake up on thyme.
My clock stopped at 8:23 a.m. I’m going to have a day of morning.
The puzzle above is based on a math trivia question I found at Trivia Cafe.
The answer could be 12:09, if the reflection in the mirror looks like this…
Then again, the answer could be 12:11, if the reflection in the mirror looks like this…
And of course, there are all the silly possibilities — for instance, if the clock is broken, it doesn’t matter what time shows in the reflection, regardless if it’s analog or digital.
Three variations of one of my favorite puzzles. The first is silly; the second is doable; and, the third will take a little bit of jiggering. I don’t know where I first saw this puzzle, but I’m pretty sure the version with ten blanks is in Gödel, Escher, Bach.
Instructions: Place numerals in the blanks to make the sentence true.
This version is for little kids. Or is it?
There are __ zeroes and __ ones in this sentence.
I’m fairly certain there are no solutions when this is extended to three blanks, but four blanks will work:
There are __ zeroes, __ ones, __ twos, and __ threes in this sentence.
It works with seven blanks (when the greatest digit is six), but that’s not much different than the one above. The piece de la resistance is the one with ten blanks:
There are __ zeroes, __ ones, __ twos, __ threes, __ fours, __ fives, __ sixes, __ sevens, __ eights, and __ nines in this sentence.
This is pretty cool. What is the value of the following expression?
(π4 + π5)1/6
This reminds me of a series of problems that I call Maximum Mileage:
- What is the maximum possible product of two positive integers whose sum is 100?
- What is the maximum possible product of two prime numbers whose sum is 100?
- A set of positive integers has a sum of 100. What is the maximum possible product of the numbers in this set?
- A set of positive numbers has a sum of 100. What is the maximum possible product of the numbers in this set?
Bonus Question: Why does the expression at the beginning of this post remind me of that series of problems?
The following sentence might be a hint:
It is fortuitous that both conundrums incur a commonality of solution.
It happened again. I received another email with a number trick that makes the ubiquitous claim, “This will work for everyone!” Sadly, it won’t, but it was kind of cool:
Calculate 39 × (your age) × 259.
The email said, “The result will surprise you.”
It didn’t. I suspected what the value of 39 × 259 would be, so I predicted the result. But if you don’t know the value of that product, then maybe you’ll be surprised.
The trick works well enough if you have a double-digit age. But my friend Ferdinand is 107 years old. His result was 1,080,807, and that just looks like a mess. The results for my six-year-old sons were better, albeit rather unsatisfying.
Ha-rumph. So much for the Internet providing mathematical inspiration.
Here are some similarly uninteresting puzzles that I created:
Calculate 7,373 × (your age) × 137.
Calculate 9,091 × (your age) × 11,111.
Calculate 101 × (your age) × 1,000,100,010,001.
To create more puzzles like this, enter factor(101010…10101) into Wolfram Alpha.
For your centenarian friends, try these:
Calculate 101,101 × (your age) × 9,901.
Calculate 3.3 × (your age) × 33.67.
Let’s not forget the little people whose age is still in the single digits:
Calculate 3 × (your age) × 37.
Calculate 41 × (your age) × 271.
And a math joke (or is it?) about age:
I’ve been good with numbers my whole life. When I turned 2, I realized that my age had doubled in one year. This concerned me… at that rate, I’d be 32 in four more years!
What goes up but never comes down?
My favorite show on TestTubeTM is Scam School, where magician Brian Blushwood takes you on a tour of bar tricks, street cons and scams. In the episode “Six the Hard Way,” he poses a mathematical challenge that is a variation on one you may have seen before. As Brian explains, “it’s almost poetic how simple this is.”
The puzzle is this: Form an expression with three 1’s, three 2’s, three 3’s, and so on, up to three 9’s, so that the value of each expression is equal to 6. As an example, an expression using three 7’s is shown below. Can you find expressions using the other numbers?
0 0 0 = 6
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 – 7 ÷ 7 = 6
8 8 8 = 6
9 9 9 = 6
You can watch Six the Hard Way, but be forewarned: at least one solution for each number is given, so you may want to solve the puzzle before viewing.
Also note that some folks have posted solutions in the comments below, so scroll at your own risk.