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Lesson 1 of 10

Algebra Play · Lesson 1 of 10

Think of a Number Tricks

“Use algebra to uncover why constant-result number tricks work and design new tricks that are guaranteed to succeed.”

Learning Objectives

• Translate a sequence of numerical instructions into algebraic expressions. • Explain why a well-designed number trick can give the same result for every starting number. • Use cancellation to identify which parts of an expression depend on the starting number. • Modify a number trick so that it produces a chosen constant result. • Test a conjecture with examples and then justify it algebraically.

Number tricks can feel mysterious because the starting number is hidden. Algebra removes that mystery. Instead of choosing one particular number, we use a letter to stand for every possible starting number at once. If the final expression no longer contains that letter, then the result does not depend on which number was chosen.

This chapter uses puzzles as a reason to do algebra. The goal is not merely to obtain an answer. We want to explain why a pattern works, decide whether it always works, and eventually design our own reliable tricks.

Turning Instructions into Algebra

Suppose the starting number is x. Every instruction changes the expression that represents the current value. If the instruction says double, x becomes 2x. If it then says add 4, the value becomes 2x + 4. Keeping this running expression is the safest way to follow the trick.

Definition
Variable

A letter such as x that represents a number whose value is not fixed while we reason about a general situation.

InstructionAlgebraic value
Think of a numberx
Double it2x
Add 42x + 4
Divide by 2x + 2
Subtract the original number2
The classic constant-result trick

Problem
A student thinks of any number, doubles it, adds 4, divides by 2, and subtracts the original number. Show that the result must be 2.

  1. 1.Let the starting number be x.
  2. 2.After doubling and adding 4, the value is 2x + 4.
  3. 3.Dividing by 2 gives (2x + 4)/2 = x + 2.
  4. 4.Subtracting the original x gives x + 2 - x = 2.
  5. 5.The variable cancels, so the result is 2 for every starting number for which the stated operations are valid.

Why the Starting Number Disappears

The crucial step is x + 2 - x. The positive x and negative x cancel because they are equal and opposite. What remains is the constant 2. This is stronger than checking the trick with 5, 12, or -3. Testing examples suggests that a trick works; algebra can show that it works for all starting values at once.

A whole family of constant-result tricksLaTeX
Here a must be non-zero. If c is chosen as a multiple of a, the final constant is especially easy to predict.
Designing a trick that always gives 3

Problem
Create a version of the original trick that always ends at 3.

  1. 1.Start with x and double it: 2x.
  2. 2.To get x + 3 after dividing by 2, the expression before division should be 2x + 6.
  3. 3.So use the steps: double the number, add 6, divide by 2, subtract the original number.
  4. 4.Algebra gives (2x + 6)/2 - x = x + 3 - x = 3.
A more complicated-looking trick

Problem
Think of a number, multiply it by 4, add 20, divide by 4, then subtract the original number. Predict and justify the result.

  1. 1.Let the starting number be x.
  2. 2.The running expressions are 4x, then 4x + 20, then x + 5.
  3. 3.Subtracting x leaves 5.
  4. 4.The multiplication and division make the starting-number part return to x, while 20 becomes 5 after division.

Designing Your Own Trick

A reliable trick is built backwards from the result you want. If you want the final answer to be 7 after subtracting the original number, you need the expression immediately before that subtraction to be x + 7. You can then choose reversible operations that create x + 7.

  1. Choose the constant result you want, such as 4.
  2. Work backwards: just before subtracting the original number, arrange for the value to be x + 4.
  3. Choose a multiplier, for example 3. Then before dividing by 3 you need 3x + 12.
  4. Turn that expression into instructions: multiply by 3, add 12, divide by 3, subtract the original number.
  5. Use algebra to verify the trick before presenting it to someone else.
Try this

Create two different tricks that both end at 6. One should use doubling and the other should use multiplication by 5. Explain why both are guaranteed to work.

When Testing Is Not Enough

If a trick works for five starting numbers, that is encouraging evidence but not a proof. A sixth number might still break it. When the same algebraic simplification applies to an arbitrary x, the reasoning covers all choices together. This shift from examples to general reasoning is one of algebra's main strengths.

Common mistake

Do not replace the original number with the current expression when an instruction says 'subtract the original number'. If the original number was x, that final subtraction is by x, even though the expression may have changed several times.

Quiz

Quick check

After the steps x → 3x → 3x + 12 → divide by 3 → subtract x, what remains?

Quick check

Why does checking ten starting numbers not prove a trick always works?

Quick check

Which addition makes 'multiply by 5, add ?, divide by 5, subtract the original' always give 7?

Quick check

In a successful constant-result trick, what usually happens to the variable at the end?

Quick check

Which expression is equal to 6 for every x?

Practice Problems

Practice Problems
  1. Show algebraically that 'triple a number, add 15, divide by 3, subtract the original number' always gives 5.
  2. Modify the original double-add-divide trick so that the final answer is 9.
  3. A trick says: multiply by 6, add 24, divide by 6, subtract the original number. Predict the answer before simplifying fully.
  4. A student claims that multiplying by 2, adding 5, dividing by 2, and subtracting the original number always gives a whole number. Is the claim true? Explain.
  5. Design your own four- or five-step trick that always gives 8, and prove it using a variable.
  6. Explain the difference between testing a trick with examples and proving it with algebra.

Let the starting number be x. The expression becomes (3x + 15)/3 - x = x + 5 - x = 5.

Key Takeaways

Key Takeaways

• A variable can represent every possible starting number in a trick. • Keep a running algebraic expression after each instruction. • A constant-result trick works because the variable part eventually cancels. • Testing examples suggests a pattern; algebra can justify the pattern generally. • To design a trick, work backwards from the constant result you want.

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Next · Lesson 2

Date-Guessing Tricks and Creating Your Own Tricks