Vol. 10 No 7 - Pi Mu Epsilon

598 PI MU EPSILON JOURNAL
PROBLEMS AND SOLUTIONS 599
V. Solution by Rex H. Wu, Brooklyn, New York.
The given equation can be written in the form
X _
J - X -
10 - y
1 '
so therefore, x = k(10- y) andy= k(x- 1), where k = 1, 2, ..., 9, or perhaps
k = lim, where m is such a digit. Eliminate y between these two equations
and solve for x to get
x=
Ilk
or x=
11m
1 + k 2 m 2 + 1
For x to be an integer, either (1 + ,tl) or (m 2 + 1) must divide 11, which is
impossible since this occurs only when k = 10 or m = 10.
Also solved by Avraham Adler, New York, NY. Charles Ashbacher, Charles Ashbacher
Technologies, Hiawatha, /A, Russell Euler and Jawad Sadek, Northwest Missouri State University,
Maryville, Mark Evans, Louisville, KY, Victor G. Feser, University of Mary, Bismarck, ND, Stephen
I. Gendler, Clarion University of Pennsylvania, Richard I. Hess, Rancho Palos Verdes, CA, Carl
Libis, University of Alabama, Tuscaloosa, HenryS. Liebennan, Waban, MA, Peter A. Lindstrom,
Batavia, NY. David E. Manes, SUNY College at Oneonta, William H. Peirce, Delray Beach, FL. H.­
J. Seiffert, Berlin, Germany, Kenneth M. Wilke, Topeka, KS, and the Proposer.
Editor's comment. Although it may seem wasteful to publish five
solutions to a rather simple problem, it is the very variety and cleverness of
the solutions that make the problem interesting.
899. [Fall 1996] Proposed by Robert C. Gebhardt, Hopatcong, New
Jersey.
Find the average number of times an ordinary six-sided die must be
tossed in order that each of its six faces comes up at least once.
I. Solution by PaulS. Bruckman, Highwood, Illinois.
Let n denote the number of times a normal die is cast in order for each
face to turn up at least once. Thus n ~ 6. Let Pn denote the probability that
exactly n casts of the die are required to fulfill this requirement. Clearly, p 6
= 6!/6 6 = 5/324. If n ~ 6, we see that Pn is the probability that exactly five
different faces have turned up in n - 1 casts and that the sixth face has come
up in the nth cast. There are 5 C 6 = 6 ways to select the first five faces that
turn up, and with such a choice, there are 5 possible outcomes for the first
cast, 5 for the next cast, and so on, for a total of 5n-t outcomes. We must,
however, subtract from this the number of ways in which only four faces
turn up in then- 1 casts, and this can happen in 5·4n-t ways, except for the
cases where only 3 faces turn up, which can happen in 10·3n-t ways;and so
on. Thus the net number of ways to have exactly 5 faces turn up in n - 1
casts is equal to
say. Therefore, Pn = A,./6n, n = 6, 7, .... It is easily verified that "};~~ Pn = 1
and also that A 5 = 720 = 6!.
We seek the expectation E of the distribution, namely E = "J;n~ npn. We
find, using Ln