Martin Gardner’s Best Math Puzzles (Part 2)
Want to learn a fun number trick? More importantly, want to learn the math behind why it works? Want to know how to figure out the arrangement of 6 marbles with only a single clue? Or the smallest number of cuts needed to turn one big cube into 27 smaller ones? Keep on reading to find out!
Jason Marshall, PhD
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Martin Gardner’s Best Math Puzzles (Part 2)
In last week’s episode, we paid tribute to the great Martin Gardner, truly a one-of-a-kind person. His prolific writing spanned a huge range of topics—most notably for us math fans, he wrote the “Mathematical Games” column in Scientific American magazine for over 25 years. And, as you can imagine, that means he amassed an impressive collection of awesome puzzles over the course of his life.
In the previous Math Dude episode, we learned about three of the puzzles that he dubbed to be amongst his best—or at least they were good enough to be included in his book My Best Mathematical and Logic Puzzles. As part of our continued celebration of the 100th anniversary of Martin Gardner’s birth, this week we’re going to take a look at a few more of his best puzzles..
Puzzle #1: The Repetitious Number
Our first puzzle this week is actually the last puzzle from Part 1 of the Martin Gardner series. Although I described how to perform this parlor-trick style math puzzle called “The Repetitious Number,” I didn’t get around to describing how and why it works. To jog your memory, here’s how Mr. Gardner describes this puzzle:
“Ask spectator A to jot down any three-digit number, and then to repeat the digits in the same order to make a six-digit number (e.g., 394,394). With your back turned so that you cannot see the number, ask A to pass the sheet of paper to spectator B, who is requested to divide the number by 7.
‘Don’t worry about the remainder,’ you tell him, ‘because there won’t be any.’ B is surprised to discover that you are right. Without telling you the result, he passes it on to spectator C, who is told to divide it by 11. Once again you state that there will be no remainder, and this also proves correct.
With your back still turned, and no knowledge whatever of the figures obtained by these computations, you direct a fourth spectator, D, to divide the last result by 13. Again the division comes out even. This final result is written on a slip of paper which is folded and handed to you. Without opening it you pass it on to spectator A.
‘Open this,’ you tell him, ‘and you will find your original three-digit number.’”
You can try this trick as many times as you’d like on your friends, and you’ll find that it works every time. But why? What’s the math behind it?
Solution: The Repetitious Number
While “The Repetitious Number” seems like magic at first, a little thought shows that it’s really not. The trick to figuring out how it works is to realize that the 6-digit number you get by writing any 3-digit number twice is the same as the 6-digit number you get when you multiply any 3-digit number by 1,001.
Do you see why that’s true? The number 1,001 is equal to 1,000 + 1. So multiplying any 3-digit number by 1,001 is the same as multiplying that 3-digit number by 1,000 and then adding the result to the 3-digit number multiplied by 1. The end result is the same as if you just wrote the 3-digit number twice.
You always get back the 3-digit number you started with.
Why does this matter? Allow me to answer this question with a question: What’s 7 x 11 x 13? If you multiply these three numbers out, you’ll see that 7 x 11 x 13 = 1,001. Aha! So this “trick” is really nothing more than first multiplying a 3-digit number by 1,001, and then successively dividing it by the three prime factors of 1,001.
In other words, the elaborate steps in this trick ultimately accomplish nothing more than multiplying by 1,001 and then dividing by 1,001—the net result is that you always, always, always get back the 3-digit number you started with. It may look like magic, but it’s really math.
Puzzle #2: Scrambled Box Tops
Another entry from Martin Gardner’s best puzzle compilation is called “Scrambled Box Tops.” Here’s how he describes it:
“Imagine that you have three boxes, one containing two black marbles, one containing two white marbles, and the third, one black marble and one white marble. The boxes were labeled for their contents—BB, WW, and BW—but someone has switched the labels so that every box is now incorrectly labeled. You are allowed to take one marble at a time out of any box, without looking inside, and by this process of sampling you are to determine the contents of all three boxes. What is the smallest number of drawings needed to do this?”
Solutions: Scrambled Box Tops
I’m happy to admit that my first guess at the solution to this puzzle was wrong, and I was initially quite surprised to find out that it could be solved by looking at only a single marble! How is that possible? After all, there are 3 boxes whose contents are unknown—how can you determine exactly what’s in each of them after seeing only half the contents of one?
As I found out, the trick here is to realize that all 3 boxes are always incorrectly labeled—it’s never the case that only 2 of them are mislabeled. Once you realize this, it’s not too tough to see how drawing a single marble can tell you everything.
For example, imagine you draw a white marble from the box labeled BW. Since this box is known to be mislabeled, the only possibility is that the other marble in it is also white (since it can’t really be the BW box)—so this is actually the WW box. The other two remaining boxes are (mis)labeled WW and BW, although we now know that they must actually be the BB and BW boxes. Since all three boxes are mislabeled, the box labeled BW cannot be the actual BW box—it must therefore actually be the BB box. And finally, the box labeled BB must really be the true BW box.
It’s kind of confusing, I know—but it makes perfect sense if you think about it. And we can deduce all of this from looking at the color of only a single marble. Isn’t logic awesome? (Yes!)
Puzzle #3: Cutting the Cube
Our final puzzle for today is dubbed “Cutting the Cube” by Mr. Gardner. He describes it as follows:
“A carpenter, working with a buzz saw, wishes to cut a wooden cube, three inches on a side, into 27 one-inch cubes. He can do this easily by making six cuts through the cube, keeping the pieces together in the cube shape. Can he reduce the number of necessary cuts by rearranging the pieces after each cut?”
What do you think? Can you come up with a way to do it?
Solution: Cutting the Cube
As it turns out, the carpenter in this puzzle has done the best that he or she can do and no rearrangement of the pieces can reduce the number of required cuts to fewer than 6. How can we say this with certainty?
No rearrangement of the pieces can reduce the number of required cuts to fewer than six.
As Mr. Gardner points out in his book, the easiest way to convince yourself that no fewer than 6 cuts are required is to think about the 1-inch cube that will be extracted from the very center of the larger cube. This particular cube is completely embedded within the larger cube and none of its eventual 6 faces even exist before any cuts are made.
As such, in order to create this 1-inch cube, we have to make 6 cuts (one for each face). And since this cube at the center requires 6 cuts, the entire process of chopping the large cube up into 27 smaller cubes must require no fewer than 6 cuts as well.
Wrap Up
Okay, that’s all the math we have time for today.
If you’ve enjoyed these puzzles, I highly recommend you pick up one or two (or ten) of Martin Gardner’s wonderful books—they’re full of mathematical fun.
And be sure to check out Scientific American’s Martin Gardner tribute page for tons of additional info about Mr. Gardner and his puzzles. While you’re there. you can also pick up a copy of the ebook, Martin Gardner: The Magic and Mystery of Numbers—it’s an awesome collection of Mr. Gardner’s most popular mathematical games and puzzles.
For more fun with math, please check out my book, The Math Dude’s Quick and Dirty Guide to Algebra. And remember to become a fan of The Math Dude on Facebook, where you’ll find lots of great math posted throughout the week. If you’re on Twitter, please follow me there, too.
Until next time, this is Jason Marshall with The Math Dude’s Quick and Dirty Tips to Make Math Easier. Thanks for reading, math fans!
Math puzzle image courtesy of Shutterstock.