# A cunning involution [closed]

You've probably seen this trick question floating around the Internet before:

If 1 = 5, 2 = 10, 3 = 15, and 4 = 20, then 5 = ?

The question is twofold:

2. What is the most elegant way to define the equivalence relationship between numbers using the logic of the trick answer? (i.e. suppose that the answer you got in part 1 was indeed the unique answer to the question, and all the numbers on the right side of the equation were the unique answers to their "equalities", how would you explain the pattern for other numbers? In particular, what would 25 equal?)

• I'll present the answer I came up with myself here so as not to bump the question: If $n$ is divisible by an even power of 5 (including not being divisible by 5 at all), then $f(n) = 5n$. Otherwise, $f(n) = n/5$. A simple solution that maximizes the occurrence of the original pattern.
– user88
Apr 16 '15 at 18:25
• I think this is too broad. You're asking for possible interpretations, and there isn't a right one, so you'll just get a big list. "Most elegant" is too vague.
– xnor
Apr 16 '15 at 20:03
• Yeah, I figured that was the case after I got two wildly differing solutions that were just as elegant as mine. I guess we'll just close this and leave it alone.
– user88
Apr 16 '15 at 20:07

How about simply defining: for all positive fractions $a$ and $b$, $a \sim b$ iff $a/b$ is a power (positive or negative) of $5$?

It's an equivalence relation, with $2 \sim 10$, $3 \sim 15$, $4 \sim 20$, and $1 \sim 5 \sim 25$.

For an involution, note that each equivalence class is of the form $\{a5^n \mid n \in \mathbb Z\}$ for some representative $1 \leq a < 5$. Then $f : \mathbb Q^+ \to \mathbb Q^+ : a5^n \to a5^{n \text{ xor } 1}$ is an involution.

• The trick question specifically states that 5 does not ~ with 25, though.
– user88
Apr 16 '15 at 20:28
• Right, I noticed that. Updated -- it is still sort of elegant, but probably only if you're a mathematician. :(
– Lynn
Apr 16 '15 at 20:34

A logical continuation of the sequence, assuming that equality takes precedence (and is unique between two numbers) is the following:

1=5
2=10
3=15
4=20
5=1
6=6
7=11
8=16
9=21
10=2
11=7
12=12
13=17
14=22
15=3
16=8
17=13
18=18
19=23
20=4
21=9
22=14
23=19
24=24

which, you will notice, completely establishes equality within the numbers 1-24, with multiples of six only equalling themselves and all others forming pairs.

Now, the next value, 25, isn't defined by this sequence directly, but the natural choice is $25=29$, which starts the whole thing over again for numbers 25-48.

• That is pretty ingenious.
– user88
Apr 16 '15 at 19:03

5=1 because 1=5

For the other numbers:

1 = 5
2 = 10
3 = 15
4 = 20
5 = 1
6 = 25
7 = 30
8 = 35
9 = 40
10 = 2
11 = 45
...
15 = 3
...

Basically, if the number $n$ isn't multiple of 5, the equivalent is $(n\ DIV\ 5)\times 20 + (n\ MOD\ 5)\times 5$. If the number is multiple of 5, the equivalent is $n/5$.

Note: this establish a bijective correspondence between a number and its transformed. $f(f(x))=x$ If it's not clear, DIV means "integer division".

• n/5 * 20 can be simplified to n*4 for simplicity's sake? Also I think it should be: (n*4) + (n+5 MOD 5) [the +5 is for 1->5] Apr 16 '15 at 15:51
• @MarkN (9 DIV 5) *20 = 1*20 = 20 while (9*4) = 36 No, that's not equivalent! Apr 16 '15 at 15:53

My guess:

5 = 1, as 1 = 5.

At least for one of the answers.

As for the relationship between the numbers:

1 = 5, 2 = 10, 3 = 15, and 4 = 20, then 5 = ?


x = 5*x so for 25, it wouldn't exist in this equation as it's cyclic, but I'm not certain yet.

• That is indeed the correct answer, but the meat of the puzzle is trying to do the second question. How do you define all numbers using this "equation pattern"?
– user88
Apr 16 '15 at 15:16
• For example, what does 25 equal, since 5 doesn't equal 25?
– user88
Apr 16 '15 at 15:18