Context: Think of a number trick: 1. Think of a number. 2. Double it. 3. Add four. 4. Divide by two. 5. Subtract the original number you thought of. Q. I predict you get 2. Am I right? Try it out with different starting numbers. Do you always end up with the same value, 2? Why?

Understand the Question A number trick can be proved for all possible starting numbers by letting the unknown number be represented by a variable, such as . By performing each operation on sequentially, we simplify the resulting algebraic expression. If the variable cancels out completely to leave a constant value , it proves the result is always regardless of the chosen number. Step 1 · Test with Numerical Examples Example 1: Starting with 1. Think of a number: 2. Double it: 3. Add four: 4. Divide by two: 5. Subtract original number: Example 2: Starting with 1. Think of a number: 2. Double it: 3. Add four: 4. Divide by two: 5. Subtract original number: Step 2 · General Proof Using Algebra Let the starting number be . 1. Think of a number: 2. Double it: 3. Add four: 4. Divide by two: 5. Subtract the original number: Since the starting variable cancels out entirely, the final result is always for any number. Answer > Yes, the prediction is correct. We always end up with because the algebraic expression simplifies to , canceling out the chosen starting number. Common Mistakes Partial Division Error: Dividing only one term by instead of the whole expression, e.g. incorrectly writing or . Sign Errors in Subtraction: Forgetting to subtract the original variable correctly at the final step.

Algebra Play | IT

Question 1

Context: Think of a number trick:

Think of a number.
Double it.
Add four.
Divide by two.
Subtract the original number you thought of.

Q. I predict you get 2. Am I right? Try it out with different starting numbers. Do you always end up with the same value, 2? Why?

Solution

Understand the Question

A number trick can be proved for all possible starting numbers by letting the unknown number be represented by a variable, such as $x$ .
By performing each operation on $x$ sequentially, we simplify the resulting algebraic expression.
If the variable $x$ cancels out completely to leave a constant value $2$ , it proves the result is always $2$ regardless of the chosen number.

Step 1 · Test with Numerical Examples

Example 1: Starting with $5$

Think of a number: $5$
Double it: $5 \times 2 = 10$
Add four: $10 + 4 = 14$
Divide by two: $14 \div 2 = 7$
Subtract original number: $7 - 5 = 2$

Example 2: Starting with $10$

Think of a number: $10$
Double it: $10 \times 2 = 20$
Add four: $20 + 4 = 24$
Divide by two: $24 \div 2 = 12$
Subtract original number: $12 - 10 = 2$

Step 2 · General Proof Using Algebra

Let the starting number be $x$ .

Think of a number: $x$
Double it: $x \times 2 = 2x$
Add four: $2x + 4$
Divide by two:

\begin{aligned} \dfrac{2x + 4}{2} &= \dfrac{2x}{2} + \dfrac{4}{2} \\[0.6em] &= x + 2 \end{aligned}

Subtract the original number:

\begin{aligned} (x + 2) - x &= x - x + 2 \\ &= 2 \end{aligned}

Since the starting variable $x$ cancels out entirely, the final result is always $2$ for any number.

Answer

Yes, the prediction is correct. We always end up with $2$ because the algebraic expression simplifies to $(x + 2) - x = 2$ , canceling out the chosen starting number.

Common Mistakes

Partial Division Error: Dividing only one term by $2$ instead of the whole expression, e.g. incorrectly writing $\dfrac{2x+4}{2} = x + 4$ or $2x + 2$ .
Sign Errors in Subtraction: Forgetting to subtract the original variable correctly at the final step.

Question 1

More questions in IT