Digits
Solution
As the last line suggests, the point of the puzzle is to find a solution to the above system of equations where all of through are digits in the set . This is a logic puzzle, and it turns out that there is a unique such way to do this besides setting all the digits to zero. It is given by letting the digits through each appear exactly once as shown below, and letting all the other letters equal :
Reading the letters gives FESTIVAL, the answer.
The expected way to break in to the logic puzzle is to start with the equation , which implies , which thus gives , and hence either or . Then work downwards from there, seeing which variables are forced to be , and considering the cases that could arise when they are nonzero. The equations are ordered in such a way that new variables are introduced sequentially, thus we usually do not need to look too many lines ahead. If we guess that no nonzero digit is repeated, this can simplify the analysis, but this assumption is not necessary to complete the logic puzzle.
Below we include a full solve-path for the logic puzzle split into cases based on whether or . In the former case , we recover the above solution; in the latter case we arrive only at the all-zero solution.
Case where
First note that implies .
Since we assume , then implies as well.
Then the equation implies either or . However, cannot be true since it would cause , contradicting the assumption that . So we must be in the case that .
Then we have This gives , which may be rewritten as Hence either or . In either case we have .
Since , it follows .
Since , it follows that so .
We now know .
We have .
Consider the equation . If , then , contradiction. So .
The next three equations should be considered at the same time: they give We consider two cases based on what we know about . If and , the first equation says , so , which is impossible.
So it must instead be the case that and ; hence we have , , and . The first equation gives . But the latter situation implies , which gives no value for .
Hence we must have and . Then , and .
Now . Hence and , thus too.
Case where
Suppose now . We prove that all the variables are zero, so this is not the desired solution.
As before implies .
Since it follows too.
From we get as well.
Since we get .
Since we get .
Since we get .
Since we get .
Since we get .
Since we get .
Since we get .
Since we get .
Since we get .
Since we get .