SAT Espionage: Cracking the Test-Makers' Code
The SAT likes to test student’s nerves by replacing numbers with letters. This can be frustrating, since most high school math problems involve at least one number. The best way to approach these problems is as a code-breaker; they have given you enough clues, if only you can find them. Here’s a game that will help your student discover that cracking the code is a lot easier than it looks.
What You Need:
- A pencil
- A few pieces of paper
- 2 or more players (you can be one.)
- Candy or another small prize
What You Do:
- Print or copy the following examples of codes: “Rhe girst oetter nf yhe qord bs trong.”“Ths sntnc hs n vwls.” “The has words scrambled been the of order” “Edoc doog a eb nac sdrawkcab gnitirw.” “ancay ouyay alktay otay igspay inpay atlinay??”
- Give each player five minutes to crack the codes just by using common sense.
- Have the players tell you their answers and compare them to the right ones: “The first letter of each word is wrong.” “This sentence has not vowels.” “The order of the words has been scrambled.” “Writing backwards can be a good code.” “Can you talk to pigs in Latin?”
- Offer a prize to the person who got the most right, or to everyone who broke more than 3 codes.
- "Extend the lesson to math. Show them this SAT problem: “If AB + BA = CDC, what does C equal? (A,B,C, and D represent separate digits within a larger number.) This looks impossible! ..." If we know neither A, B, or D, how can we know C? This is where we need to use a little bit of common sense to crack the code. We could spend a lot of time plugging in numbers for A and B, and trying to come up with an answer that fits the form of CDC. Or, we could use some critical thinking. What’s the LARGEST number you can create by adding two 2-digit numbers? If we add 99 and 99, we’ll get 198. We can never get a number bigger than that; we’ll never even get to 200. No matter what A and B are, the number must be in the hundreds. Therefore, the only possible value for C is “1.”
- One more problem: 9 xJ KL If K = 4, then L = ? There are two ways to do this: First, we can try to think of a number in the forties that is a multiple of nine. The only possibility is 45, so L must be 5. Or, we might remember that the digits of any multiple of 9 add up to 9. Therefore, we know that K + L = 9, so L = 5.
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