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## Algebra...Scam School Style

Published on Thursday, September 01, 2011 in , , , , , , ,

Any type of math can inspire dread in a student, but algebra can cause confusion even students who have grasped all the prior math.

Why not make it fun? Try performing an algebra-based magic trick, and then having your students work through the process with algebra to see why it works. Strangely enough, Scam School is a perfect resource for this.

This week's episode of Scam School, featuring an amazing card coincidence, is a good place to start. Let's take a look:

At about 9 minutes into the video, notice that both one of the ladies and Brian Brushwood remark that they have no idea how it works (I imagine Brian probably does, but thought it was a better idea to keep everyone more at ease). The reason it works, of course, is algebra. It's time to take a closer look at the math.

The first mathematically significant step is the division of the 21 matches. All 21 matches must be divided between 2 people, and both people must have at least 1. So one person will have A amount of matches, which is at least 1 and no more than 20, and the other person has 21 - A matches.

Take a closer look at what this means. The smallest amount of matches either person can have, as already stated, is 1, with the other person having 20. What happens if we increase the number of matches from 1? With 2 matches, the other person will have 19, and so on up to 10, with the other person having 11 matches. As soon as we give the person with 10 matches 1 more match to make 11, then the other person winds up with 10!

Stated more simply, there will always be one person with a number of matches ranging from 1 to 10. We can't say who this is, but that doesn't matter any more than not being able to say how many matches each person is holding.

Naturally, at the end of the trick, you want the people to have 2 different cards, so how do you prevent them from having the same number? This is insured by the odd number of matches (21 in this video), and goes back to the old rule of an even number plus an odd number always being an odd number. If one person has an odd number of matches, then the other person must have an even number of matches for them to add up to 21. Further, since one person has an odd number of matches and the other even, they must have a different number of matches!

When this number is applied to choosing cards, that means they must be choosing cards at 2 different positions. Speaking of cards, it's time to turn our attention to them.

At this point, we know that one person is thinking of a card in position 1 through 10, and the other is thinking of a number at a position from 11 through 20. What happens when you remove those top 10 cards? Let's consider just the removal of the 10 cards first, without the reversal of the cards.

The removed packet, of course, contains one person's card, while the other person's card is still in the talon (remaining majority of the card deck). The position of the card which is still in the talon has, of course, been reduced by 10 cards. If it was the 11th card, it's now 11 - 10 = 1st card. If it's 20th, it's now the 10th card of the talon, and so on. Note that 11 through 20 is a total of 10 cards.

Let's take a look at this mathematically, defining A as the person who winds up with 1-10 matches:

$\\(1 < A < 10) \\ A + B = 21 \\ A + B - A = 21 - A \\ B = 21 - A \\ (remove \ 10 \ cards \ at \ this \ point) \\ B = 21 - 10 - A \\ B = 11 - A$

So, we know A is in a packet of 10 cards somewhere, and we know B is in a different packet of 10 cards (well, 10 cards plus the remainder of the deck, but we know no selected cards are there). Thanks to the above formula, we also see that they share a definite relationship: B = 11 - A.

Think about this pattern. If one person's card is in position 1, the other person's card is at position 10 in the other pile. If one person's card is in position 2, the other person's is in position 9, and so on. In other words, wherever one person's card is, the other person's card is in the reverse position!

If either of the piles is reversed, it's not too hard to see that both cards would be in the same position in both piles! That's why exactly 10 cards need to be reversed in the trick, and why the trick itself works!

This is hardly scam school's only algebra trick, as well. If you understood the math here, try working out some of Scam School's other algebraic challenges, including Purloined Objects, Reading 5 Minds at Once (James Grime can help you work this one out), Mathemagic (I'll help you with this one), and The Trick That Fooled Einstein (this one practically GIVES you the algebraic equation!).

These are hardly the only Scam School episodes with algebra-based tricks, but they are the best places to start for getting interesting lesson ideas. Algebra is about finding patterns, even when you don't know the numbers involved. Therefore, the real trick is to get students interested in examining the pattern in the first place!

### 2 Response to Algebra...Scam School Style

Anonymous
8:05 AM

There's a typo. Even number + odd number is always an odd number, not even :)

2:16 AM

Thanks for the catch!

I just updated it.