Algebra Play · Lesson 2 of 10
Date-Guessing Tricks and Creating Your Own Tricks
“Use variables and place value to explain, decode, and redesign a date-guessing algebra trick.”
• Represent a month and day with separate variables and follow a multi-step date trick. • Explain why multiplying the month information by 100 separates it from the day. • Decode a hidden date from the final output of the trick. • Check whether a decoded pair is a valid calendar date. • Modify constants in the trick and create a new version that can still be decoded.
A date-guessing trick looks more impressive than a one-number trick because it hides two pieces of information at once: the month and the day. Algebra shows that the trick is really an encoding system. The operations rearrange the month and day so that they can later be recovered.
Encoding the Month and Day
Let M stand for the month number and D stand for the day. For 26 January, M = 1 and D = 26. For 25 December, M = 12 and D = 25. We keep M and D separate while following the instructions.
| Step | Expression |
|---|---|
| Start with month | M |
| Multiply by 5 | 5M |
| Add 6 | 5M + 6 |
| Multiply by 4 | 20M + 24 |
| Add 9 | 20M + 33 |
| Multiply by 5 | 100M + 165 |
| Add the day | 100M + 165 + D |
The complicated-looking middle steps have one purpose: they turn M into 100M and create the fixed extra amount 165. Adding D at the end produces 100M + 165 + D.
Problem
Follow the trick for 26 January.
- 1.Here M = 1 and D = 26.
- 2.Use F = 100M + 165 + D.
- 3.F = 100(1) + 165 + 26 = 291.
- 4.The same value is obtained by performing each instruction one at a time.
Decoding the Final Number
To undo the fixed part of the trick, subtract 165. This leaves 100M + D. Because a day is at most 31, D fits in the final two decimal places. The digits before those last two places identify the month.
Problem
A player reports 1390. Which date was chosen?
- 1.Subtract the fixed amount: 1390 - 165 = 1225.
- 2.Read 1225 as 100M + D.
- 3.The last two digits give D = 25.
- 4.The remaining digits give M = 12.
- 5.The date is 25 December.
Problem
A player reports 296. Find the date.
- 1.Subtract 165: 296 - 165 = 131.
- 2.Interpret 131 as 100M + D, so M = 1 and D = 31.
- 3.The date is 31 January.
- 4.Do not require the decoded number to have four digits; January dates naturally produce values such as 101, 109, or 131 after subtracting 165.
Why Place Value Makes the Trick Work
Multiplying M by 100 shifts the month information two places to the left. That leaves room for a two-digit day. This is the same place-value idea used when writing a two-digit number as 10a + b or a three-digit number as 100a + 10b + c.
Representing information in a form that follows a rule so that the original information can later be recovered.
The restriction D ≤ 31 is essential. If the second quantity could be 147, then 100M + D would no longer separate cleanly into a month part and a two-digit part. The trick succeeds because the range of possible days is known in advance.
Algebra may decode a pair such as M = 2 and D = 31, but that pair is not a valid date. After decoding, check that the month lies from 1 to 12 and that the day is possible for that month.
Changing the Trick Without Losing the Date
We can change the middle operations as long as we know the fixed amount they create and still finish with a coefficient of 100 on M before adding D. For instance, multiply M by 10, add 7, multiply by 10, then add D. The result is 100M + 70 + D. Subtracting 70 recovers 100M + D.
Problem
Create a shorter date trick whose final value is 100M + 240 + D.
- 1.Start with M and multiply by 10 to get 10M.
- 2.Add 24 to get 10M + 24.
- 3.Multiply by 10 to get 100M + 240.
- 4.Add D.
- 5.To decode, subtract 240, then read the month and day from 100M + D.
- Choose a fixed constant C that you can reproduce through your steps.
- Make sure the coefficient of M becomes 100 before D is added.
- Finish with a form 100M + C + D.
- Tell the decoder to subtract C.
- Test January, a two-digit month such as October, and a date near the end of a month.
Use the original rule to decode final values 1269 and 394. Then decide whether both decoded dates are valid.
Quiz
In the original date trick, what remains after subtracting 165 from the final value?
Why is the coefficient 100 useful?
If F = 1269 in the original trick, what is F - 165?
Which decoded pair must be rejected as an invalid date?
A new trick gives F = 100M + 70 + D. What should be subtracted first when decoding?
Practice Problems
- Decode the original trick when the final value is 1269.
- Decode the original trick when the final value is 394 and identify the date.
- Show that 25 December produces 1390 without performing every instruction separately.
- Create a date trick of the form 100M + 90 + D and write clear instructions for a player.
- A student designs a rule that finishes at 10M + D. Explain why this is not reliable for all dates.
- Invent a different constant C, create a valid encoding 100M + C + D, and explain how a friend would decode it.
1269 - 165 = 1104, so M = 11 and D = 4. The date is 4 November.
Key Takeaways
• A date trick can encode two quantities by using place value. • The original steps simplify to F = 100M + 165 + D. • Subtracting 165 leaves 100M + D, so the last two digits can represent the day. • Decoded values must still be checked against real calendar constraints. • A new trick remains decodable if it produces a known form 100M + C + D.