# Byzantine Basketball

The problem of *Byzantine Basketball* with solution was sent to me by Prof. W. McWorter:

Byzantine Basketball is like regular basketball except that foul shots are worth *a* points instead of two points and field shots are worth *b* points instead of three points. Moreover, in Byzantine Basketball there are exactly 35 scores that never occur in a game, one of which is 58. What are *a* and *b*?

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

Byzantine Basketball is like regular basketball except that foul shots are worth *a* points instead of two points and field shots are worth *b* points instead of three points. Moreover, in Byzantine Basketball there are exactly 35 scores that never occur in a game, one of which is 58. What are *a* and *b*?

### Solution

Without difficulty one can see that, for there to be only finitely many scores that never occur, *a* and *b* must be relatively prime positive integers. All scores have the form *a*x + *b*y, where x and y are nonnegative integers. Let's say a nonnegative integer *c* is *representable* if *c* has the form *a*x + *b*y, with x and y nonnegative integers, and *c* *nonrepresentable* otherwise. We count the number of nonrepresentable integers. The plan is to show that (*a* - 1)(*b* - 1) - 1 is the largest nonrepresentable nonnegative integer, and that exactly half the integers in the interval [0, (*a* - 1)(*b* - 1) - 1] comprise all nonrepresentable nonnegative integers.

First we show that all integers greater than (*a* - 1)(*b* - 1) - 1 are representable. Any such integer can be written (*a*-1)(*b*-1)-1 + *c*, for some *c* > 0. Since *a* and *b* are relatively prime, *c* can be written *c* = *a*u + *b*v, where -*b* < u ≤ 0 and v > 0 (see Remark 1.) Thus,

(*a* - 1)(*b* - 1) - 1 + *c* = *a*(*b* - 1) - *b* + *c* = *a*(*b* - 1) - *b* + *a*u + *b*v = *a*(*b* - 1 + u) + *b*(-1 + v).

The restrictions on u and v insure that *b* - 1 + u ≥ 0 and -1 + v ≥ 0. Hence (*a* - 1)(*b* - 1) - 1 + *c* is representable, and so all integers greater than (*a* - 1)(*b* - 1) - 1 are representable.

Next, we show that (*a* - 1)(*b* - 1) - 1 is not representable. Suppose that (*a* - 1)(*b* - 1) - 1 = *a*x + *b*y, for some nonnegative x and y. Then

*a*x + *b*y = (*a* - 1)(*b* - 1) - 1 = *a*(*b* - 1) - *b*.

Hence *a*(*b* - 1 - x) = *b*(y + 1). Since x ≥ 0, and y ≥ 0, we have y + 1 > 0 and *b* - 1 - x is positive and less than *b*. Hence, since *a* and *b* are relatively prime, *b* divides *b* - 1 - x, a contradiction. Thus (*a* - 1)(*b* - 1) - 1 is not representable, and, with the previous paragraph, it is the largest nonrepresentable integer.

Finally, we show that exactly half the integers between 0 and (*a* - 1)(*b* - 1) - 1 are nonrepresentable. We show this by showing that the map f(*c*) = (*a* - 1)(*b* - 1) - 1 - *c* interchanges representable integers with nonrepresentable integers (see Remark 2.) If *c* is representable and f(*c*) is representable, then so is (*a* - 1)(*b* - 1) - 1 = *c* + f(*c*), contradiction. Hence f(*c*) is nonrepresentable if *c* is representable. Now suppose *c* is nonrepresentable. Then, since *a* and *b* are relatively prime, there are integers u and v such that *c* = *a*u + *b*v, with 0 ≤ u < *b* (see Remark 1) and v < 0 (*c* nonrepresentable forces

f(*c*) = (*a* - 1)(*b* - 1) - 1 - *c* = *a*(*b* - 1) - *b**a*u - *b*v = *a*(*b* - 1 - u) + *b*(-1 - v).

The restrictions on u and v imply that *b* - 1 - u ≥ 0 and *c*) is representable.

Well, that just about does it. Now we know that there are exactly *a* - 1)(*b* - 1)/2*a* - 1)(*b* - 1)/2 = 35.*a* = 2,*b* = 71;*a* = 3,*b* = 36;*a* = 6,*b* = 15;*a* = 8,*b* = 11.*b* are not relatively prime. *a* = 2*b* = 71*a* = 8*b* = 11,

### Remark 1

c = *a*x + *b*y, for some integers x and y, then *c* = *a*(x + *b*t) + *b*(y - *a*t), for every integer t. Hence, for some t, x + *b*t is between 0 and *b* - 1, and, for some other t, x + *b*t is between -*b* + 1 and 0. In short, there is always a choice for x which is a residue modulo *b*.

### Remark 2

The function f maps [0, (*a* - 1)(*b* - 1) - 1] onto itself.

### Remark 3

The problem was included into the Putnam 1971 competition. Two of its solutions appear in *Mathematical Gems, II*, by Ross Honsberger (MAA, 1976).

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

64959006 |