# An Identifiably Very Uncomplex Puzzle

This is a relatively easy puzzle, but I thought I might share it anyhow.

If 5 is 4, 6 is 9, and 7 is 5, what is 8?

Hint 1:

The title gives a clue (not directly to the answer, but to the general rule.)

Hint 2:

SPQR up!

• Title has identity so perhaps it's $1$? Jul 9, 2018 at 18:44
• Edited Hint 1 for clarity.
– akr
Jul 9, 2018 at 18:46

The answer is that 8 =

1

because

we are looking at Roman numerals embedded in the written-out form of the number's name. fIVe=IV=4, sIX=IX=9, seVen=V=5. So eIght=I=1.

– akr
Jul 9, 2018 at 18:48

I'll give it a shot:

8 is 25.

Reasoning:

'five' contains 4 letters
'six' contains 3 letters -> squared is 9
'seven' contains 5 letters
So basically just how many letters, and if the number is even, square it.

• Not quite my reasoning, but close! The same, simple rule applies to all of them.
– akr
Jul 9, 2018 at 18:39

You could also have that

$8=8$

Explanation:

$5$ has two lines and one curve. If the curve means 'subtract one line' then the result is $5$ minus $2-1$ which is $4$.

Similarly, $7$ has two lines. The result is $7$ minus $2$ which is $5$.

So the rule for odd number is *odd number - (number of lines - number of curves)

For the even numbers, it could just be that they are flipped upside-down, in which case $6$ is $9$ and $8$ remains the same.

• Not quite! But good try!
– akr
Jul 9, 2018 at 18:40