13.2 Thinking About ‘Think of a Number’ Tricks
In Grade 7 you ran into “think of a number” tricks. Try this one with whatever number you fancy.
- Think of a number.
- Triple it.
- Add twelve.
- Divide by three.
- Take away the number you first thought of.
My guess is you finish on 4. Did I get it right? Test it with 7, with 40, with \tfrac12, the answer is 4 every single time. How can a trick name your result before you’ve even worked it out?
The trick to the trick is to walk through the steps not with one fixed number but with a letter-number x that stands for whatever you picked.
Worked Example
Why the trick always lands on 4.
| Step | In words | In algebra |
|---|---|---|
| 1 | Think of a number | x |
| 2 | Triple it | 3x |
| 3 | Add twelve | 3x + 12 |
| 4 | Divide by three | \dfrac{3x+12}{3} = x + 4 |
| 5 | Take away the original | (x+4) - x = 4 |
The x wipes itself out at the final step, so the answer is 4 no matter what x was. That is the entire secret.
The Key Idea A number trick “works for everybody” precisely when the chosen number cancels out of the closing expression. Algebra lets us watch that cancellation unfold, and lets us arrange it deliberately.
Math Talk How would you tweak the steps so the answer always comes out 6? What about 10? (Hint: here “+12” turned into “÷3 → +4”. Which number should you add at step 3 to be left with 6?) Can you string together a longer set of steps that still funnels everyone to one fixed value?
Use the checker below to type in your own steps and watch whether the chosen number cancels.
Worked Example
Design a trick that always finishes on 6.
Work backwards. We want the closing step “subtract the original x” to be applied to something of the form x + 6, since (x+6) - x = 6. To reach x + 6 after a “divide by three”, the line before division must read 3x + 18 (because \tfrac{3x+18}{3} = x + 6). So:
| Step | In words | In algebra |
|---|---|---|
| 1 | Think of a number | x |
| 2 | Triple it | 3x |
| 3 | Add eighteen | 3x + 18 |
| 4 | Divide by three | x + 6 |
| 5 | Subtract the original | 6 |
Replacing “add twelve” by “add eighteen” shifts the answer from 4 to 6. Inventing a trick is just running the algebra in reverse.
A trick that finds a date
Here is a livelier one. Aarav asks Diya to picture a date and keep it to herself. She settles on Independence Day, 15 August, written as day 15, month 8.
| Step | Instruction | Diya’s working |
|---|---|---|
| 1 | Think of a date | 15/08 |
| 2 | Multiply the month by 5 | 8\times 5 = 40 |
| 3 | Add 6 | 40 + 6 = 46 |
| 4 | Multiply by 4 | 46 \times 4 = 184 |
| 5 | Add 9 | 184 + 9 = 193 |
| 6 | Multiply by 5 | 193 \times 5 = 965 |
| 7 | Add the day | 965 + 15 = 980 |
Diya announces 980, and Aarav shoots straight back, “You were thinking of 15 August!” How does he do it?
Let the month be M and the day be D, and run the same steps in algebra:
\begin{aligned} 5M \;\xrightarrow{+6}\; 5M+6 \;&\xrightarrow{\times 4}\; 20M+24 \;\xrightarrow{+9}\; 20M+33\\ &\xrightarrow{\times 5}\; 100M+165 \;\xrightarrow{+D}\; 100M + 165 + D. \end{aligned}
So the final answer is always 100M + 165 + D. Aarav simply subtracts 165:
980 - 165 = 815 = 100M + D.
Since the day D is never more than 31, only two digits, the last two digits of 815 are the day and whatever sits before them is the month. Here M = 8 and D = 15: the 15th of August.
Worked Example
Diya’s next answer is 1395. Which date?
Subtract 165: \;1395 - 165 = 1230 = 100M + D. The last two digits give D = 30, and what remains is M = 12. The date is the 30th of December.
Worked Example
A friend reports 379. Find the date, and the catch.
Subtract 165: \;379 - 165 = 214 = 100M + D. The last two digits give D = 14, and the rest is M = 2. The date is 14 February. Notice the answer 379 is only three digits because the month is single-digit; the recipe still works because the “100M” part keeps the day safely parked in the last two places.
Try This Recover the date behind each final answer. (i) 1170 (ii) 487 (iii) 996
Then test the trick on a friend, asking them to start from their birthday.
Subtract 165 from each answer; the last two digits are the day D, and the rest is the month M.
- 1170 - 165 = 1005 \Rightarrow M = 10,\ D = 05: 5 October.
- \;487 - 165 = 322 \Rightarrow M = 3,\ D = 22: 22 March.
- \;996 - 165 = 831 \Rightarrow M = 8,\ D = 31: 31 August.
Math Talk Can you reshuffle the steps and still get the date back? If you change step 3 or step 5, the number you subtract at the end may no longer be 165. Work out what it should become, that calculation is the act of inventing your own trick.