# My Mathematical Magic Show: Part 8c

This mathematical trick was not part of my Pi Day magic show but probably should have been… I’ve performed this for my Precalculus classes in the past but flat forgot about it when organizing my Pi Day show. The next time I perform a magic show, I’ll do this one right after the 1089 trick. (I think I learned this trick from a Martin Gardner book when I was young, but I’m not sure about that.)

Here’s a description of the trick. I give my audience a deck of cards and ask them to select six cards between ace and nine (in other words, no tens, jacks, queens, or kings). The card are placed face up, side by side.

After about 5-10 seconds, I secretly write a pull out a card from the deck and place it face down above the others. I then announce that we’re going to some addition together… with the understanding that I’ll never write down a number larger than 9. For example, the 4 and 6 of spades are next to each other. Obviously, $4+6 = 10$, but my rule is that I’m going to write down a number larger than 9. So I’ll subtract 9 whenever necessary: $10-9 = 1$. Since 1 corresponds to ace, I place an ace about the 4 and 6 of spades.

Continuing in this way (and having the audience participate in the arithmetic so that this doesn’t get boring), I eventually get to this position: Finally, I add the two cards at the top (and, in this case, subtract 9) to get $6+9-9 = 6$, and I dramatically turn over the last card to reveal a 6. I’ll often perform this trick when teaching Precalculus, as the final answer involving Pascal’s triangle. As discussed yesterday, suppose that the six cards are $a$, $b$, $c$, $d$, $e$, and $f$. Forgetting for now about subtracting by 9, here’s how the triangle unfolds (turning the triangle upside down): $a \qquad \qquad \qquad \quad b \qquad \qquad \qquad \quad c \qquad \qquad \qquad \quad d \qquad \qquad \qquad \quad e \qquad \qquad \qquad \quad f$ $a+b \qquad \qquad \qquad b+c \qquad \qquad \qquad c+d \qquad \qquad \qquad d+e \qquad \qquad \qquad e+f$ $a+2b+c \qquad \qquad b+2c+d \qquad \qquad c+2d+e \qquad \qquad d+2e+f$ $a+3b+3c+d \qquad \quad b+3c+3d+e \qquad \quad c+3d+3e+f$ $a+4b+6c+4d+e \qquad b+4c+6d+5e+f$ $a+5b+10c+10d+5e+f$

Not surprisingly, the coefficients in the above chart involve the numbers in Pascal’s triangle. Indeed, the reason that I chose to use 6 cards (as opposed to any other number of cards) is that the bottom row has only 1, 5, and 10 as coefficients, and $10 \equiv 1 (\mod 9)$. Therefore, the only tricky part of the calculation is multiplying $b+e$ by $5$, as the final answer can then be found by adding the remaining four numbers.

My students usually find this to be a clever application of Pascal’s triangle for impressing their friends after class. P.S. After typing this series, it hit me that it’s really easy to do this trick mod 10 (which means getting rids of only the face cards prior to the trick). All the magician has to do is subtly ensure that the second and fifth cards are both even or both odd, so that $b+e$ is even and hence $5(b+e)$ is a multiple of 10. Therefore, since $10c+10d$ is also a multiple of 10, the answer will be just $a+f$ or $a+f-10$.

(If the magician can’t control the placement of the second and fifth cards so that one is even and one is odd, the answer will be just $a+f+5$ or $a+f-5$.)

Henceforth, I’ll be doing this trick mod 10 instead of mod 9.

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