This is not what you may think or what you would like to think. It is a play on “The root of passion is kiss”. Is it now clear? No? Well how about now?
This is about finding the values of the letters assuming that each letter has a unique single digit value and the triviality of P and K being zero is not considered. How would one go about doing this? Many of us would not attempt it simply because it would take too long. The programming kind would quickly write a program and get the job done. The real thrill would lie in a paper and pen approach. Ideally till you get an answer or at least till you reach a point where only enumeration may help. To begin with we have 9999 four digit numbers that can be cut down to size.
Now PASSION is a 7 digit number and the minimum and the maximum values for PASSION are 1022345 and 9877654. The square root of the maximum and minimum are 3142.x and 1011.x. Since letter K , I and S represent three different numbers the maximum will now be 3122 and the minimum will be 1022.
Since S*S is ending in N , S cannot be 1,5 or 6 since a square of number ending in 1 will also end in 1 and same holds true for 5 and 6. Since K is not equal to S, the (K, S) pairs will be (1,2),(1,3),(1,4),(1,7),(1,8). It will not be (1,9) because then N will be 1 and N is not equal to P. The (K, S) pairs will also include (2,3),(2,4),(2.7),(2.8),(2.9),(3,1),(3,2). This implies that there are a maximum of 96 values of KISS (8 into 12).
Now ‘I’ cannot be a zero. The reasoning is as follows. If ‘I’ is 0 then KISS is the of the form (1000K+SS)^2.
This means that on expansion kit will be 1000000 or 4000000 or 9000000 + whatever SS squared is. Now SS squared will never more than 9801. Hence the total will be of the form (1,4,9)00NNNN where NNNN is SS squared and the first digit is one of 1,4 or 9. Now A is not equal to S hence ‘I’ is not equal to 0.
This further implies that there are a maximum of 84 values of KISS (7 into 12). Since ‘I’ is not equal to N this 84 possible value can be further reduced to 60 by removing numbers of the type 2933 since that would make N as 9 and N is not equal to I.
One more refinement that is possible is to take the square root of 1988765 which is 1410+ so we can eliminate KISS up to 1410 because then P would be equal to K equal to one which is not possible.
Now that we have reduced it to a manageable numbers, time to enumerate and see the results, which those of you who are curious can find out. Have a good weekend!