The Logic Course Egg Hunt 🥚

A puzzle hidden inside the logic course. Solve it, and you could win.

How it works

The logic course has 13 worksheets, and every one of them hides an Easter egg question — the slightly unserious one, with a joke buried inside a perfectly rigorous problem. Part of the fun is spotting which question it is.

12 of those Easter eggs are multiple choice, and the correct answer to each gives you a clue letter. Together, the twelve clue letters form the basis of a Morse code, and decoding it reveals the treasure word — a mathematics-related word of five distinct letters.

So to clarify:

Work through the logic course and find the Easter egg question on each worksheet.
Solve the 12 multiple-choice ones. These must be done in the order they appear in the Logic Course; each one gives you a clue letter.
Decode the treasure word. The 12 ordered clue letters, joined together, form a Morse code, which you then need to decode to reveal the treasure word — a mathematics-related word of five distinct letters.
Enter the treasure word, along with a valid email address.
If your treasure word is correct, then you will be entered into the raffle.

Already worked through the worksheets?

If you finished the logic course before this hunt existed, you'll need to download the latest version of the worksheets — the Easter egg questions, and the clue letters they carry, only live on the updated ones.

The prize

One winner takes home their choice of:

🎓 2 hours of one-to-one tuition with me — the headline prize, and worth comfortably more than the alternative below; or
🎁 a £100 Amazon voucher, if you'd prefer.

Timeline

Entries close: 31st August 2026
Correct entrants notified: by 1st September 2026
Winning number determined: Saturday 5th September 2026, by the National Lottery Lotto draw
Winner announced: 6th September 2026

The draw is a genuine lottery drawn from the correct entries: every correct entry has an equal chance of winning. I'll reveal the exact method before entries close on the 31st of August 2026. Whatever it is, it will be fair — and it won't be possible for me to fix the result, if that's a concern. I already have a few ideas; if you'd like to suggest one, write to me at jzmaths@hotmail.com.

The rules (the short, sensible ones)

You must enter a valid email address. No valid email, no entry — that's how we reach you if you win.
One entry per person. Suspicious multiple entries may be queried, and I reserve the right to void them.
Winners are drawn only from correct entries. The raffle rewards solving it, not being fastest — a late correct answer has exactly the same chance as an early one.
If the winner is under 18, we'll need a parent or guardian's consent to arrange the tuition or send the voucher.

One more thing

This isn't really about the prize. It's an excuse to spend some time with logic — implication, negation, contradiction, the little traps that catch everyone the first time — presented as a hunt rather than a lesson. I hope you enjoy it. Best of luck. 🍀

How the prize winner will be selected (update)

The prize winner will be selected using the official UK National Lottery Lotto draw on Saturday 5 September 2026. This gives us a public source of randomness which is entirely outside my control.

Settling the list of entries

There is nothing you need to do between entering and the draw. Every correct entry is automatically included, and you do not need to reply to anything to remain in the draw.

Duplicate entries — including several entries submitted within a few minutes of one another from the same network address — will be reduced to a single entry. I am going to be nice about this: rather than disqualifying every entry involved, I will keep the earliest valid entry and remove the rest. Good try, though 🙂

Once the final list has been settled, let $M$ be the number of correct entries. Each entry will be assigned a number from $0$ to $M-1$ , in the order in which the entries were originally submitted. This ordering has no effect on anyone's chance of winning.

When entries close on 31 August, the final list and the value of $M$ will be locked; neither can change afterwards. On 1 September, everyone with a correct entry will be emailed both their assigned number and the value of $M$ . Every participant will therefore know the total size of the draw and their own place in it before the lottery numbers are drawn.

Using the National Lottery results

The Saturday 5 September Lotto draw will produce two rounds, each containing six main numbers and a Bonus Ball. Only the twelve main numbers will be used; the Bonus Balls will be ignored.

The six main numbers from Round 1 will be written first, in the order shown on the official National Lottery results page, followed by the six main numbers from Round 2. Numbers within each round will be separated by hyphens, with a vertical bar separating the two rounds. There will be no spaces or leading zeros.

For example, if the results were:

Round 1: 11, 28, 41, 48, 57, 59
Round 2: 7, 8, 20, 21, 22, 52

the resulting string would be:

11-28-41-48-57-59|7-8-20-21-22-52

I will then calculate the SHA-256 hash of that exact string and interpret the resulting 64 hexadecimal digits as a single non-negative integer $N$ . SHA-256 is deterministic, so anyone using the same string will obtain exactly the same value of $N$ .

You can verify the calculation yourself using free online tools: search for a SHA-256 generator to calculate the hash, followed by a hexadecimal-to-decimal converter to convert the hash into $N$ .

The winner will be the participant whose assigned number is the remainder when $N$ is divided by $M$ . Dividing by $M$ always produces a remainder from $0$ to $M-1$ , so the result will correspond to exactly one participant. Since $N$ cannot be known until the lottery balls have been drawn, each of the $M$ assigned numbers is, for all practical purposes, equally likely to win.

For the example above, the SHA-256 hash is:

01cded0558268e5e2e4bcea75b40826233cd88402928b8e014116c7b1332739e

Interpreting those 64 hexadecimal digits as an ordinary decimal integer gives the following 75-digit number:

816152354830100611821330557368477464889221606955577130964219875915136660382

This is $N$ . If there were 50 correct entries, dividing $N$ by 50 would leave a remainder of 32, so the participant assigned number 32 would win.

After the draw, I will publish on this page the exact input string, its SHA-256 hash, the decimal value of $N$ , the final value of $M$ , the remainder and the winning assigned number. This will allow anyone to reproduce the complete calculation independently and verify the result.

If the winner does not reply

The winner will have 5 days from the day I first contact them to reply and claim the prize. If they do not reply within that time, the prize will pass to the participant holding the previous assigned number: the winning number minus one, and $0$ goes to $M-1$ .

That participant will then have 5 days of their own to reply and claim the prize. This process will continue, moving backwards through the assigned numbers and wrapping around where necessary, until the prize is claimed.

It has ended!

Thank you to everyone who participated in the egg hunt — I hope you learnt some logic along the way!

While writing the logic course worksheets, I started slipping in Easter egg questions: unserious, but logically rigorous and hopefully funny. I love setting them and cannot seem to help myself — what started as an itch has developed into a ritual! 😄

To celebrate finishing the course, I decided to turn their answer keys into a coded puzzle. I rearranged the answer options so that the correct answers, read in order, gave BBABBAAAAAAA. With A representing a dot and B a dash, this becomes the unspaced Morse code for MATHS: -- .- - .... ....

Unspaced Morse code is usually ambiguous, so I gave the hint that the treasure word was mathematics-related and contained five different letters, to help you find the intended word.

I imagine some of you got BBABBAAAAAAA and simply fed it into ChatGPT. 😂 And that’s absolutely fine — working through the worksheets and finding the correct answers was the main part: learning some logic and having a bit of fun!

I’m pleased to say there were 65 correct entries, so M = 65, and I have now emailed everyone their assigned number. If you submitted MATHS (upper or lower case, both are considered correct), but have not received an email, please contact me at jzmaths@hotmail.com and I’ll check.

Good luck for the draw on Saturday!

The result

At the time of writing, I have found these results on YouTube and across several news outlets. The official National Lottery website has not yet published them, but I am all but certain that they are correct.

The draw took place on Saturday 5 September 2026, using the official National Lottery Lotto results and exactly the method published above.

The two rounds produced:

Round 1: 3, 4, 10, 20, 49, 51 — Bonus Ball 36
Round 2: 8, 21, 26, 28, 35, 36 — Bonus Ball 52

As stated above, only the twelve main numbers are used and both Bonus Balls are ignored. Writing Round 1 first, then Round 2, gives the input string

3-4-10-20-49-51|8-21-26-28-35-36

whose SHA-256 hash is

dc16beaed7e913fab76e0351591b428a6f57aea38061a280dd6f0e039d797357

Reading those 64 hexadecimal digits as a single decimal integer gives $N$ :

99549013369318365643512798913847675710456570582253263434341252329805566276439

There were $M=65$ correct entries, numbered $0$ to $64$ . Dividing gives

N \equiv 29 \pmod{65}

🏆 The winning number is 29

Many congratulations to the holder of number 29! I will be in touch to arrange the prize once the results have been confirmed on the official National Lottery website.

Thank you again to everyone who took part. I hope you enjoyed hunting through the worksheets, learnt some more logic along the way, and had a little fun decoding MATHS. It was wonderful to receive 65 correct entries—and although there can only be one prize winner, every one of you successfully completed the hunt!

Best of luck to everyone with your TMUA preparation!

Perhaps I will have to hide a few more Easter eggs in future... 😄