# Thread: tricky question

1. ## tricky question

Hi can anyone please help to solve the below problem?

If
5-3=4
2-8=2
5-1=6
6-3=3
1-7=?

please advise as how this is solved methodically

Keshav

2. ## Re: tricky question

I don't believe there is a unique answer to the problem as written. For example we could assume that the equations you provided use symbols 1, 2, 3, 4, 5, 6, 7, and 8 which are assigned to values as follows:

$\begin{matrix}Symbol & Value \\ 1 & 1 \\ 2 & anything \\ 3 & 2 \\ 4 & 3 \\ 5 & 5 \\ 6 & 4 \\ 8 & 0 \end{matrix}$

This mapping satisfies all the equations given. For example the first equation with symbols 5-3 has a values of 5-2, or 3, which is equivalent to symbol 4. But note that the symbol 2 can be any value you want, since any value minus zero equals that value (per the second equation). You have no infomation as to what the symbol 7 represents - it could be 6, or 600, or anything at all. Hence the value of symbols 1-7 could be anything. So I'm wondering - are there more conditions provided?