Find the number n such that the following alphanumeric equation:
KYOTO
KYOTO
+ KYOTO
TOKYO
has a solution in the base-n number system.
(Each letter in the equation denotes a digit in this system, and different letters denote different digits)
I have read over the previous comments but I don't think the explanations given were thorough enough. Here I show the reasoning I used to find the solution.
I'll number the columns in the summation from 1 to 5 in order, 1 being the leftmost and 5 the rightmost. The obvious place to begin looking at the problem is column 5. We are adding 3 Os together and the digit at the bottom of the column is also an O. Now, it IS possible that we may have to perform a carry from column 5 into column 4. (NOTE - as we are adding only 3 numbers in this problem, all carry digits will be either 0, 1 or 2)
So O+O+O=kn+O, where n is the base we are looking for and k is the carry digit (0, 1 or 2). Hence
3O=kn+O
2O=kn
O=(k/2)n
If k=0, then O=0 also. If k=1, then O=n/2, that is the digit O is equal to half the base (this means also that in this case the base would have to be an EVEN number). If k=2, O equals n but this is not possible for a digit in base n so we disregard this case. So, thus far we have proven that the digit O is either equal to zero, or is equal to half of the base, where the base is some even number.
Now let's consider the second of these two cases. That is, let's assume that O is not 0; therefore our base is even and O equals half the base (O=n/2). Now, consider column 3. If we didn't carry anything into column 3 from column 4, then the digit at the bottom of column 3 would be an O, just as it is in column 5. This means that we carried either a 1 or a 2 into column 3 from column 4. So, O+O+O+c=kn+K, where c is the digit carried INTO column 3 (either 1 or 2) and k is the digit we carry OUT OF column 3.
O+O+O+c=kn+K
(n/2)+(n/2)+O+c=kn+K
O+c=(k-1)n+K
So, O plus c (1 or 2) equals digit K plus (perhaps) some amount to be carried out of column 3.
At this point let's look back at column 1; here we see that K+K+K (plus perhaps a carry digit of 1 or 2) equals T. Hence K must be less than n/3, for otherwise we would have to carry a digit out of the first column, and that won't fit the given problem.
Now, we know that the base n is even; also, it must be at least 4 as there are four letters in the problem and we are told that they all represent DIFFERENT digits. If the base n was 6, O (=n/2) would be 3 and then O plus c (1 or 2) would be 4 or 5, hence K would be 4 or 5 which would make K too large to work in column 1. A similar problem occurs here for any even base greater than 4; these cases will always result in K being too large. But what about the case when the base is equal to 4? If n=4, then O=2, and if the carry digit c coming into column 3 were to equal 2, we would have K=0; I'll allow the reader to convince himself that there is no way to effect a solution to the problem under these circumstances though (you can't get the remaining digits 1 and 3 to fit the remaining letters T and Y in a way to make the sum work out).
So we assumed that O was not equal to 0 and this led us to a situation where there was no solution to the problem. Now we'll have to go back and look at the case where O=0.
(At this point I will post this comment as part 1 of a full solution - I'm aware that sometimes these message boards can only take posts of a certain size and I don't want to get cut off! The rest of the solution will follow in part 2 which I will post shortly)
-John
Edited on September 21, 2003, 3:19 pm
Edited on September 21, 2003, 4:00 pm
No Subject
