So the thing with math is that sometimes it feels like magic with how it all works. The kick is that once you understand it and spring it on someone in a way that’s easy to understand, you can amaze people with how cool math really is. Number tricks are some of the easiest feats to pull to amaze the youngins out there.
Trick 1: Short and Simple
The first one is one I somehow remember since reading it in a seventh grade textbook. Our school was just switching to IMPACT Mathematics at the time and I got a taste of the Course 2 book. Something that I remember was all the various skits to open sections in there, and one always just stood out to me. I’ll cut out the interspersed dialogue from the skit and instead give you the “trick”:
“Think of a number. Double it and add 20. Now halve your answer. Subtract the number you first thought of.”
So. What number did you arrive at? Write it down, then try it again with a different number and write down your result. Keep trying. Can you escape what’s happening?
Let us try one together. Perhaps a wacky number like 43. Let us double it which gives us 86. Adding 20 gives us 106. Halving it gives us 53. Then subtracting 43 gives us 10. That number just keeps showing up, doesn’t it?
So why not look at “x”. “x” is arbitrary here. Doubling “x” gives us “2x”. Add 20? So “2x+20”. Halving that is “x+10”. Subtract the original number, which is “x”. That leaves 10. Looks like we can’t escape our fate. No matter what happens, we will ALWAYS arrive at 10.
Trick 2: Card Magic
This next one is designed to be done with cards. Let us examine what lies on those cards:
| 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 |
| 17 | 19 | 21 | 23 | 25 | 27 | 29 | 31 |
| 2 | 3 | 6 | 7 | 10 | 11 | 14 | 15 |
| 18 | 19 | 22 | 23 | 26 | 27 | 30 | 31 |
| 4 | 5 | 6 | 7 | 10 | 11 | 12 | 13 |
| 20 | 21 | 22 | 23 | 28 | 29 | 30 | 31 |
| 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 |
| 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 |
| 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 |
So think of a number between 1 and 31. Which cards are they on? I’d like for you to take note of what the lowest number on those cards are and list them out, in order, next to the number you first thought of. Try this out with a few numbers, then see if you can pick up on what the “trick” is.
I won’t get into details on how it works, because I think this is one that’s best for you to learn. Then, see if you can come up with variations of it, like cards that have numbers 1-63, maybe. Or 1-15. Perhaps, if you’re crazy, go for 1-127. If you’re insane, see what happens when you try to make cards with different “lowest numbers”. Have fun!
Trick 3: Three Digit Swapping
This last one is a little more complex, and this is the one I hinted at in the last post about square numbers. I learned this trick from what was then known as The Math Circle at Harvard University (now The Global Math Circle), which I went to for a portion of fifth grade based on good word from our neighbors.
So think of a number in the hundreds, but make sure that the digits in the hundreds and ones places are different. Let’s use 274, for example. Swap the digits in the ones and hundreds place, then subtract whichever number is smaller from the one that is larger. For 274, the “swapped” number is 472, so we’ll do 472 – 274. After determining the difference, do a digit swap again, but add instead of subtracting. From our example, the difference is 198. The new “swapped” number is 891, and the sum is 1089. That number seems familiar, doesn’t it?
Try this out with a few numbers. Much like the first trick, it seems like it’s fated to always have a sum of 1089. Let’s try this with arbitrary digits:
| A | B | C | |
| – | C | B | A |
| ? | ? | ? |
We’ll have to rely on our knowledge of subtraction here to guess what the quotient is. Because we know that we had to take the number that was larger than the smaller to be subtracted from (i.e. 472 is larger, thus, we are subtracting from 472 rather than 274), we know that A is bigger than C. However, because we must start from C – A, we have to essentially “borrow” every time, and it leads to the 9 below the B’s.
| A | B | C | |
| – | C | B | A |
| ? | 9 | ? |
So it now begs the question of what happens with how we define “10 + C – A”, which is what is actually happening with the ones digits. Because “A” and “C” are arbitrary, there isn’t much we can do in terms of defining that difference, so we set an arbitrary variable “E”, while the result of (A-1) – C will be another variable “D”. This leaves our difference, for the time being, as “D9E” (do not read this in another base/radix).
| A | B | C | |
| – | C | B | A |
| D | 9 | E |
What happens afterward? Well, we take “D9E” and add “E9D”. So now we have to effectively add “D” and “E” together:
| D | 9 | E | ||
| + | E | 9 | D | |
| ? | ? | ? | ? |
So let us recall how we defined “D” and “E”. D = A – 1 – C, while E = 10+ C – A. Add these two quantities together and we have: A – 1 – C + (10 + C – A) = A – A – 1 + 10 – C + C = 9. So that means that D + E = E + D = 9. So when we take into account that we also have a 9 + 9, we must add a 1 to the D + E, which is 10. So in the end, we have:
| D | 9 | E | ||
| + | E | 9 | D | |
| 1 | 0 | 8 | 9 |
Crazy, isn’t it? Analyzing the trick is so interesting that I often revisit it. In the process, I’ve also found some additional intricacies with some of the numbers involved in the process, for example, “D9E” (and “E9D”) as a result. What are the possible numbers that satisfy the pattern? What is it that you notice about these numbers? I leave that for you to figure out and how to possibly prove your hypothesis.
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