Well, that calculator looked like an ordinary one when I saw it in the shop, but it behaves most extraordinarily! I tried using the + and * keys on small integers, but the results I got were totally wrong and usually decidedly non-integer. However, there does seem to be some method in its madness. Its results are never negative and always fairly small. Provided that it doesn't display Error -- which it does in many cases, unfortunately.

Whatever these operations are that the + and * keys do, they behave in some ways like $+$ and $*$. If I denote them by $\oplus$ and $\otimes$, then, for any $a, b, c$:

$a\oplus b=b\oplus a \\ (a\oplus b)\oplus c=a\oplus(b\oplus c) \\ a\otimes b=b\otimes a \\ (a\otimes b)\otimes c=a\otimes(b\otimes c) \\ a\otimes (b\oplus c)=(a\otimes b)\oplus(a\otimes c)$

That is, $\oplus$ and $\otimes$ are commutative and associative, and $\otimes$ is distributive over $\oplus$.

$10$ has this simple property: for any $a$, $10\otimes a=a$.

Hint 1:

Each other number $a$, as entered on the keypad or displayed, has an internal equivalent $f(a)$. Whenever $f(a)+f(b)=f(c)$, $a\oplus b=c$, and whenever $f(a)*f(b)=f(c)$, $a\otimes b=c$. The image or range of $f$ is a field.

I also found this curious relationship between $\oplus$ and $\otimes$:

$a\otimes b=s\oplus d$

where $s$ and $d$ are the sum and difference of $a$ and $b$, found, I hasten to add, by using ordinary arithmetic, not by using the + key!

Occasionally it produces an integer. What integers are these values? In each case, give the lowest possible.

  1. $13\oplus 7$
  2. $z$ where, for all $a$, $a\oplus z=a$ and $z\otimes a=z$
  3. $24\otimes 14\otimes 7\otimes 6$
  4. $14\otimes 6\otimes 4\otimes 3$
  5. $13\otimes 12\otimes 11$
  6. $\sqrt{10\oplus 10\oplus 10}$

And what's the hint in them taken collectively?

  • $\begingroup$ 'Its results are never negative and always fairly small', '10⊗a=a'; Aren't these statements contradictory to each other? Or would it error when a is negative? $\endgroup$
    – Arth
    Jun 16, 2016 at 12:04
  • $\begingroup$ Try using &times; for the multiply symbol. I wonder if this SE supports the key format? <kbd>+</kbd>? $\endgroup$
    – JDługosz
    Jun 16, 2016 at 18:16

2 Answers 2


I have an answer that works. I haven't yet proved it to be unique. We can take

$f(x) = 2 \cos (\frac{\pi}{30}x)$.

Then the mysterious sum-and-difference identity turns into

a standard trigonometrical identity.

The magic number

30 is there to make $f(10)=1$ as required.

The answers to the requested calculations are, in order,

3, 15, 19, 9, 14, 5

which are the alphabet-positions of the letters of


I think the statement in hint 1 is not quite correct, by the way:

we have f(x)=f(-x) for any x and f(x+60)=f(x), so f doesn't have a well-defined inverse, which means that it can't possibly be true that e.g. "whenever $f(a)+f(b)=f(c)$, $a\oplus b=c$" -- because whenever the stated condition holds for a,b,c it also holds for a,b,-c and a,b,c+60 and so on, and $a\oplus b=c$ can't be true for more than one choice of c.


A start:


This is equivalent to 10⊗3, as 13 is the sum of 10 and 3, and 7 the difference. For any a, 10⊗a = a, so the answer to this is 3


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