by Victor Bryant
Published Friday December 29 2023
Here are some clues about my PIN:
- It has 3 digits (or it might be 4).
- The first digit is 3 (or it might be 4).
- The last digit is 3 (or it might be 4).
- It is divisible by 3 (or it might be 4).
- It differs from a square by 3 (or it might be 4).
In each of those statements above just one of the guesses is true, and the lower guess is true in 3 cases (or it might be 4).
That should enable you to get down to 3 (or it might be 4) possibilities for my PIN. But, even if you could choose any one of the statements, knowing which guess was correct in that statement would not enable you to determine the PIN.
What is my PIN?
Sunday Times Teasers Discussion Group 
WELL SET vICTOR
Remarkable Teaser, but straightforward if laborious to solve.
It was nice to have a numerical rather than a geometrical question (I am aware I skipped 3191 because of time and not seeing a shortcut but plan to go back to it).
Not too difficult. There aren’t an unreasonable number of squares starting 3 or 4 and a difference of 3 or 4 ending 3 or 4 narrows the possibilities to not many. Then look at the 3 result. The 3 or 4 cases narrows the list down to 3 or 4 as Victor says. The conditional statement determines the 3 or 4 for each requirement, resulting in the one PIN.
Now it is nearly time to dance the night away.
I started at the other end. Discounted 303, 304…493, 494.
Started again 3003, 3004…4993, 4994 and picked out multiples of 3. 4 but not 12.
The square test got me down to five candidates, but the last two logic steps took more thinking.
An analysis of all of the possible squares is one way to solve this teaser. However, the use of modulo arithmetic leads to a simple synthesis of the 3 (or it might be 4) possibilities for the PIN. The answer soon follows.
I agree it was a well set logic problem which I found easier by considering the limited number of applicable squares to each range of numbers (303-494, 3003 to 4994), in drawing up a Truth Table ,to eventually get 3 (or 4?) candidates, before applying the constraints 2-liner.
After realising PIN=444 was not allowed it resolved quickly to viable PIN options- although final 2-line constraint was tricky to deal with, I thought.
Like Ben C., for the final steps I used a truth table, which eventually made sense.
I considered how possible squares could lead to possible PINs.
If a PIN ends in 4, one can show that it must be divisible by 4. That places a very tight constraint on possible PINs.
If the PIN ends in 3, one can show that if the square is n^2, then n = 0 (mod 6).
The final paragraph rapidly eliminates all but one of the possible PINs.