Note: Sorry, this puzzle almost similar with the last question:

The main purpose of this QUESTION is:

  • To know what types of this puzzle is exactly.
  • It is still can be solve if I remove the most two important hint(A*D=A) and (A=8)?




Hint 1 A*D=A

Hint 2 A=8


The answer is HUMAN (8,21,13,1,14)after mapping .And represent to A,B,C,D,E before mapping . As you can see, in this Example I'm using simple characters and numbers mapping.


These type of question is usually referred to as a system of simultaneous equations.
To start with (without referring to the hints), you have five unknowns $A, B, C, D, E$ and three equations.

Such a system, where there are more unknowns than equations describing them are usually referred to as underdetermined. This means that you probably don't have enough information to determine a unique solution. In other words, if a solution exists, then there will be an infinite number of solutions.

In this case, you really just have two equations, which I'll write as $$ B = A+C $$ $$E = C+D $$ The third equation is automatically satisfied if the first two are.

For example, I can pick any numbers $x$, $y$ and $z$ I wish and set
$A = x$,   $B = x+y$,   $C = y$,   $D = z$,   $E = y+z$
and the first three equations will always be satisfied.

Now, if we include the hints, this restricts the system a bit more. In particular, $$ A = 8, AD=A \Rightarrow x = 8, z = 1$$ Then, the overall solution is determined by one parameter, $y$ and we have $$ (A,B,C,D,E) = (8, y+8, y, 1, y+1) $$ The HUMAN solution you suggested comes from picking $y=13$ but I could also, for example, pick $y=7$ to get $$ (A,B,C,D,E) = (8, 15, 7, 1, 8)$$ which also satisfies all of the equations and spells HOGAH.
Since there is just one free parameter left, you just need to have one more equation to guarantee your solution is unique. For example, you could say $B + C + E = 48$

  • 2
    $\begingroup$ In the same way that cryptograms don't always specify the unknown shift (eg rot13, rot7 etc), I like how the unknown value of c makes this question more interesting. $\endgroup$ – Caleb May 21 '18 at 17:07

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