Question

Number system unknown. Find the number n such that the alphanumeric equation KYOTO + KYOTO + KYOTO = TOKYO has a solution in the base-n number system. (As usual, each letter in the equation denotes a digit in this system, and different letters denote different digits.)

Answer 1

n=9 to give:
K=1, Y=3, O=0 and T=4 all in base 9 number system

Answer 2

How did you get that? This question drove me crazy!

Answer 3

OK, I first re-wrote it as follows:
KYOTO
KYOTO
KYOTO
------
TOKYO

Then I reasoned that since we have 4 distinct letters, the minimum base must be 4.
Then, looking at the rightmost column, I said to myself that the only way 3 * O can give you a number that is O plus some possible carry is for O to be zero.

I then rewrote the problem replacing all O's with zeros as:
KY0T0
KY0T0
KY0T0
------
T0KY0

I then said, looking at the 2nd column from the right, that 3T must give a number that involves a carry (of 1 or 2) since the next column to the left contains all zeros but its corresponding digit in the answer is \(K\). We therefore get:\[3T=Y+n\quad\text{or}\quad3T=Y+2n\]and since the carry over is 1 or 2, this gives:\[K=1\quad\text{or}\quad K=2\]which means there is no carry over to the next column to the left which contains all \(Y's\).
Now, since the total of the \(Y\) column contains a zero in the corresponding digit position in the answer, we can say:\[3Y=n\quad\text{or}\quad3Y=2n\]Now, if \(3Y=n\), then \(n\) must be a multiple of 3.
So I tried base 6 first which gives:\[
Y=2\quad\implies3T=Y+n=2+6=8\quad\text{which can be rejected as 8 is not a multiple of 3}\]Then I tried base 9 first which gives:\[
Y=3\quad\implies3T=Y+n=3+9=12\quad\implies T=4
\]
So, using \(Y=3,T=4\) we can rewrite the problem in base 6 as:
K3040
K3040
K3040
------
40K30

which implies \(K=1\).

I hope my explaination was clear.

Answer 4

Very, thanks!