[Guest]

Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy this alphametic equation. None of R and T can be zero.

(REKNIT)/(TINKER) = (RE - 1)/(RE)

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

[Guest]

Moving the equation around we can get:

(REKNIT) / (TINKER) = (RE - 1) / (RE)

(RE) * (REKNIT) = (RE) * (TINKER) - (TINKER)

(RE) * ((TINKER) - (REKNIT)) = (TINKER)

(TINKER) - (REKNIT) = (TINKER) / (RE)

Let's define x = (TINKER) - (REKNIT)

We can determine that x is going to be at most five digits. From the subtraction side of the equation, we can conclude that T = R + 1, because there is no other way to achieve a five digit result.

Using T = R + 1, we can conclude that the last digit of the result of the subtraction problem is 9. This means that the product of 9 and E ends with R. This narrows our choices for (E, R) to (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1). We can conclude that the combination (5, 5) does not work because we know that the missing digits do not repeat. Furthermore, R cannot be 9 because T would have to be 10 or 0, neither which are possible solutions.

Using these combinations, we can determine that the first digit of x is 1. In each combination, we know R and E, we can thus determine T in each group. Assume the biggest possible number by filling in uncertain digits with 9, we find that no matter which combination the result is between 10,000 and 19,999.

Given that the first digit of x is 1, we can conclude that 10 + I - E = 1 or 10 + I - 1 - E = 1 depending on whether or not the one is carried over from the previous digit in the subtraction problem. This means that (I, E) = (1, 9), (0, 9) or (0, 8). The solution (1, 9) does not work because both I and R would equal 1. Using (0, 8), we can determine using the subtraction part of the equation that the last two digits of x would be 79. The product of 79 and RE, which is 28 using this solution set, does not result in [ER], or 82. Using TI--ER = 20--91 does work though. The difference between 91 and 02 is 89. The product of 89 and 19 ends in 91. Moreover, x is now equal to 1--89.

Let us substitute a and b as the two digits in x. Multiplying x by 19 should give us 20NK91. Converting digits to actual variables gives us:

19 * (10089 + 1000 * a + 100 * b) = 200091 + 1000 * N + 100 * K.

This equation can be boiled down to: 19 * (10a + b) = 84 + 10 * N + K.

This shows us that 10a + b is equal to 05, 06, 07, 08 or 09. The corresponding N + K are 11, 30, 49, 68, and 87. The first three answers do not work because of repeating digits.

Plugging in the last two, we find that only 10 * N + K = 87 works.

Thus, (TINKER) = 208791.

Edited July 23, 2009 by jchang320