Vol. 5 No 10 - Pi Mu Epsilon

JOURNAL

ELEMENTARY NUMBER THEORY IN CERTAINSUBSETS OF THE INTEGERS, IPI MU EPSILON JOURNALTHE OFFICIAL PUBLICATIONOF THE HONORARY MATHEMATICAL FRATERNITYDavid C. Kay, EditorASSOCIATE EDITORSRoy B. DealLeon BankoffOFFICERS OF THE FRATERNITYPresident: H. T. Karnes, Louisiana State UniversityVice-president: E. Allon Davis, University of UtahSecretary-Treasurer: R. V. Andree, University of OklahomaPast-President: J. C. Eaves, University of West VirginiaCOUNCILORS:E. Maurice Beesley, University of NevadaGloria C. Hewitt, University of MontanaDale W. Lick, Russell Sage CollegeEileen L. Poiani, St. Peter's CollegeChapter reports, books for review, problems for solution and solutions to problems,should be mailed directly to the special editors found in this issue underthe various sections. Editorial correspondence; including manuscripts and newsitems should be mailed to THE EDITOR OF THE PI MU EPSILON JOURNAL,601 Elm, Room 423, The University of Oklahoma, Norman, Oklahoma 73069.Far manuscripts, authors are requested to identify themselves as to their class oryear if they are undergraduates or graduates, and the college or university theyare attending, and as to position if they are faculty members or in a non-academicprofession.PI MU EPSILON JOURNAL is published semi-annually at The University ofOklahoma.SUBSCRIPTION PRICE: To individual members, $4.00 for 2 years; to nonmembersand libraries, $6.00 for 2 years; all back numbers, $6.00 for 4 issues,$2.00 per single issue; Subscriptions, orders for back numbers and correspondenceconcerning subscriptions and advertising should be addressed to the PI MU"SILON JOURNAL, 601 Elm Avenue, Room 423, The University of Oklahoma,rman. Oklahoma 73069.BIJ Cme.n 0. Adiio and Julian R. KoLodThe. Cotie-ge- 06 Saint ROACThe purpose of thi paper is to study some aspects ^f elementarynumber theory in certain subsets of ':, the set of integers.The moti-vation for this study came about from discussions in number theory textsindicating that the fundamental therrem of arithmetic does not hold insome ubsets of Z (see [l, p 281, [2, p 181, or [4, p 121). The usualexample given is 22, the multiples if 2, where "t is hown that 2, 6,and 18 are "prime" and that 36 = 6 6 = 2 - 18.In elemental", textbooks on algebra, divisibility i- usually rtudiedin an inregral domain (commu~arive ring with uniry and no zero divisors)and, as is well known, the fundamental theorem does nut always hi Id int h i ~ case eyther.The definitirn. are usually -tated for a commutativerizg and the main results usually hold in an integral domain.:n thispaper we make no such algebraic assumptions en the -ub et: con 'dered.The assumptions which are made are more or less forced by The facr Thai-we need TO limit he kinds of subsers of Z we wish ro consyder bur areweak enough ti allow a wide variety uf sets tv be taken into dnridcra-tion. However, our ma!n re-ults w'll be for the ring nZ, the multiplesof some integer n, n1. These results, however, do not hold becausecf the algebraic properties of nZ but because the primes; and i:omposites(as we shall define them) are so nicely distributed in these sets.The author? would like to thank the referee for hi? observationsand comments.1. PhtLw-inOJU.e,fiIn looking at nZ one immediately sees that this subset of Z isclosed under negation.To discuss divisibility in subsets of Z, thisis precisely the restriction we wish to place on the types of subsetswe are to consider (with the exception that we exclude the singletonset (0) ).Thus in this paper we only consider non-empty subsets A ofZ having the two properties: (a) A if (0) and (b) if x A, then -x f- A.

Tc deflna divisibility in such subsets of Z we merely mimic theusual definition.Ve.&iviLtion: Let x, y c A, x # 0.We say that x divides y in A(or x A-divides y) if there exists n e A such that y = nx.To denote the fact that x A-divides y we write x(A)y; otherwise wewrite ~ ( 4 ) ~ If . A = Z, we adopt the usual notations x \ y and x 1[ y.As a result of this definition the following can be easily verified:(1) If 1 e A, then l(A)x for all x in A.(2) Divisibility in A is reflexive if and only if 1 A.(3) x(A)(-x) if and only if -x(A)x if and only if -1 e A.(4) If 0 A, then 0(/)x for all x in A, but x(A)0 for all non-zero x in A.(5) If A = B and x(A)y, then x(B)y. Thus, if x(A)y, then x 1 y.On the orher hand, if 0 is the set of odd integers, thenx(0)y if and only if x 1 y.(6) Divisibility in A need not be transitive. For example, takeA = { 22, '3, 26, *is}. Here 2(A)6 and 6(A)18, but 2(/).18.Although divisibility in A need not be transitive, about the bestwe can do in this regard is rhe following proposirion, which is easilyverified.Lemma 1': If A is closed under multiplication, then divisibilityin A is transitive.,As a result, divisibility in )zZ is transitive.this proposition is false as we see later.The converse ofTo define primes in A we again mimic the usual definition.-.Ve-dinLti-on: Let p i- A. p # 1 is said to be prime in A, (or to bean -A-prirne )- ,(1) if 1 i A, then x(/)~ for all x in A or(2) if 1 6 A and x(A)p, then x = *p or x = *l.X is composite in A if x is nor \5Thus, for example, 2, 6, 10, 14;--or *l and is not ~rl-e in Aare prixe ir. 22 and the primesin Z are prime in any set containing them. More generally, if p A c Band P is prime in B, then p is prime in A.The converse is false--take4 i- 42 c 22. Note also that p is prime in A if and only if -p is primein A, so as a result we shall omit discussing the negative primes.There are only two subsets of Z which do not have primes, namely{*!I and { 0, *l}. For any other set A, the least positive integer inA which is greater than one is always prime in A.It is also easy tosee that for any integer n > 1, there is a subset of Z with -n and n asits only primes.in Z, the setFinally, if pl, p2,---p are any n distinct positivenhas exactly 2n primes (n positive primes).Note that divisibility istransitive in A so that A serves as a counterexample to the converseof Lemma 1.2. The. Fundwntd Tl~eo~tem 0(1 WlundcIn this section we discuss the primes in nZ, the fundamental theo-rem of arithmetic in nZ and make some observations concerning primes innZ and Z.For the remainder of the paper we use FTA1 to mean thereexists a factorization into a product of primes, FTA2 to mean the fac-torization is unique, and FTA to mean both FTA1 and FTA2.The FTA1 always holds in a subset A of Z of the type consideredearlier (other than A = { *l} and A = [O, el}) and the proof can be pat-terned after that given in [4, p 111. We shall show, however, thatFTA2 does not hold in nZ.FTA does hold; for example take A = {any (positive) prime in Z.There are proper subsets of Z for which the- - ,-p2, -p, p, p2, ' " ' } where p isThere is a simple formula for the primes in n2'(n > 1) while, ofcourse, no such formula i-5 known to exist for the primes in Z.Thiofiem 1':The positive primes in nZ (n > 1) are of the form(kn + i)n for l = 1,2,.;n - 1 and k = 0,1,2;--. All other posiriveintegers in nZ are composite and are of the form kn 2 for k = 1,2;".Proof: Let p c nZ so that p = mn for some rn Z, and suppose thatp is prime in nZ.rn = kn + i, 0 s i < n.Using the division algorithm (in Z) we can writeNowi = 0 iff p = kn 2 iff n(nZ)p denying the

primality of p. Thus if p is an nZ-prime, then p = (kn + ¥i) fori = 1,2,- - -, n - 1.ThiOJLem 2: For any integer n 3 2, the FTA2 does not hold in n2.Proof: By Theorem 1, iz2(n + I ) ~ is composite in nZ and is notuniquely factorable sincen2(n + I ) ~ = n . [n(n + 1l2] andn2(n + I ) ~ = [n(n + l)] [n(n + l)]are two differenr prime facrorizations in nZ.The following rheorem identifies which composires in nZ are and whichare not uniquely factorable .in nZ.First we observe that if x is com-posite in nZ, we have x = kn7 by Theorem 1. We then note that x can bewritten as x = kiz2 = k$where m 2 2 and n 1 k,.mT h e m 3: Let x be an 7x2-composite written x = kn where m  2and n 1 k.prime in Z.Proof:Then x is uniquely factorable if and only if k = 1 or k isLet k = pla1p2a2. . .ppap be the unique prime factorizationof k in 2. Observing that k = 1 is a trivial case, we go on to thecases when k # 1.Case 1: r =1.m- 1(a) If a, = 1 then x is uniquely factorable as x = n (np).(b) If a, > 2, then x is not uniquely factorable sincem-1x = 71 (nplal )are two different prime factorizations of x in n2.cd~e2: P a 2.Then x is not uniquely factorable since- -- m-2* x = n (nplal )(np2'2. .p:r)= nm-l(nplal.. .ppa')are two different prime factorizations of x in nZ.Theorem 1 allows us to make the following obser-;2:~0:.~primes in nZ and in 2; some of these will be disc.:sse=concerningi: -.ore de~ail inthe next section.(1) There are infinitely many primes in each ?&Z, as there are inz.(2) There are not arbitrarily large gaps between primes in nZ asthere are in Z. In fact, the gap between primes in È iseither 0 or 1.(3) There are infinitely many twin primes (primes separated byonly one composite) in each nZ.Moreover, the twin primes innZ have the form: (kn + (n - 1)) n and ((k + DM + 1) . nfor k = 0,1,2, .-.(4) Given any positive integer n, however large, nZ contains ?a - 1consecutive primes.primes is the sequence 2, 3.)3. The AflA,t/vn&tic Function n [ XI(In Z the largest sequence of consecutiveL ~ TT (x) denote rhe number of posirive primes in A which are less thanAor equal to x e A, Since the primes and composites in iiZ are so nicelydistributed, we are able to obtain an explicit formula for this function.In the following theorem, [x] means the greatest integer (in Z) which isless than or equal to x.Theokem 4:If x = kn e. nZ, thenn - 1(kn) = k- [k/n] =-nz(kr ) + r ,where in the last equality k = nq + r and 0 2 r < 72, q = [k/x].Proof',The number of positive primes in nZ which are less than orequal to kn is equal to T - C whereandT = the total number of p6sftive integers in ?22 which are .< kn,C = the number of positive composites in ni! which are  kn.It is clear that T = k.are of the form kn2,As we saw in Theorem 1, the composites in nZNow the composites ip nZ which are less than orequal to kn are : n2, 2n2, 0 * , [k/iz] n2. Thus, C = [k/n]-In Z, the ratio n(x)/x can be interpreted in two ways':(1) as the number of positive primes which are  x compared to x, or(2) as the number of positive primes which are 5 x compared tothe number of positive integers which are < x.

Of course these two meanings are the same, and it is well-known thatlim v(x)/x = 0 ._C-È-cHowever, in nZ the above meanings are different since the first gives theratio IT nz(kn)/kn and the second gives the ratio n(kn)/k. In thenZnext theorem we examine the two limits arising from these ratios.The.on.ew 5: For a fixed n > 1TT nZ -(1) lim- --kk-Proof: Let q = [k/n]. Fromnz knl i m - =k-m kTheorem 4l i m +k-xaThe proof of (2) follows from (1).T ^(fen)nz(kn)We observe that for a fixed n, 0 < iim kn < l i m < 1k-xa k*and that for large n, lim vnz(kn)/kn is near zero while Iim n (kiz)/kn2k-r-k-xais near cne.2. The. AluJLIma^cc Fundore4 T [x) and o 1x1" In this section we discuss the following two functions:~ ( x = ) the number of positive divisors of x in A ,o (x) = the sum of the positive divisors of x In A.AIn 2, one normally establishes formulas for these fu~?l,:.n- by firsttaking x as prime, then x as a power of a prime, and :.-.enshows the abovefunctions are multiplicative [f(mn) = f(m) f(1: .-¥he 7 and n relarivelyprime] so that one can rely on FTA2 to obtain the formulas for composites.Since factorization is not unique in nZ, this process will not work.the following approach we see that these functions are "nearly" multipli-cative in nZ.We first establish the formulas for powers of n and thenfor composites. Clearly, if x is prime in nZ, then T (x) = 0 = a (x).nZnZWe saw before that if x is composite .In EZ, we can write x = kn m wherem  2 and n 1 k.m-1mLemma 2: If x = n m , then ~ ~ ) = ~ m - 1 ( and n o nZ (nm) = nr.r=1Proof: The positive divisors of nm in nZ are n, n< n 3 ,m-1' - ', nand the formulas follow easily.Tke.okem 6 : If x = kn m where m  2 and n ) k, thenm~ ( k ) n = (m - 1) ~ ( k = ) T (nm) ~.(k)nZProof: Each divisor of k multiplied by n is a divisor of x. Eachdivisor of k multiplied by n2 is a divisor of x. Inductively, each divisorof k multiplied by n m- 1 is a divisor of x. Since k does nothave n as a factor neither do the divisors of k and so the divisorsnamed above are all distinct.visors of x. Since each row contributes TMoreover, these are all the positive di-z(k) divisors of x and sincethere are m - 1 rows, there are therefore (m - 1) ~,,(k) divisors of xLiin nZ. To find aZ(knm) we merely add up the divisors in each row. Thesum cf the jth row is n3dle equality.oZ(k). Summing the m - 1 rows gives the mid-Lemma 2 gives the last equalities in the two formulas.In a future paper we shall discuss in nZ the notions of greatestcommon divisor, relative primeness, and Euler's function ifi(x).REFERENCES1. Andrews, '3. E., Number Theory, W. B. Saunders Company, Philadelphia,Pennsylvania, 1971.2. Barnett, I. A., Elements cf Number Theory, Prindle, Weber, andSchmidt, Inc., Bosten, Massachusetts, 1969.3. Hardy, G. H. and Wright, E. M., An Introduction to the Theory ofNumbers, Oxford 'Jniversity Press, 4th ed., New York, 1960,4. Niven, I. and Zuckerman, H. S., An Introduction to the Theory ofNumbers, 2nd ed., John Wiley and Sons, Inc. , New York, 1966.In

~A NESTED PRIME NUMBERMAGIC SQUAREReverend Victor Feser of St. Ambrose Church, St. Louis, has pointedout an example of a 13 x 13 nested magic square consisting entirely ofprime numbers.^-(See in this connection the article "1-lagic SquaresWithin Magic Squares" by Joseph Moser, this Journal, 5, No. 8 (Spring1973), p. 430.) Each smaller square centered at 5437 is also a magic square,lbom the Recreational Mathematics Journal, No. 5 (October 1961), p. 28.REFEREES FOR THIS ISSUEThe editorial staff sends a note of appreciation to the followingpersons who freely gave of their time to evaluate papers submitted forpublication prior to this issue : Joseph Lehner, University of P-i rtsburgh ;Walter Growney, Susquehanna University; and members of the mathematicsdepartment at the University of Oklahoma, R. V. Andree, Patrick Cassady,Bradford Grain, John Green, Andy Ilagid, Lee Pound (computer science 1,W. 1'. Reid, and Kirby Smith.The Journal also acknowleges with gratitude the typist For thisissue, Theresa Killgore, who is a senior at the University of 'Jklahoma.NICENESS OF THE SOCLE AND A CHARACTERIZATIONOF GROUPS OF BOUNDED ORDERby S. U. TcLU.ey1UutVin Kentnckq Univm'LtqA classification result (that tells when or how two algebraic sys-tems are of the same kind) is one of the most desired results in thestudy of any algebraic system.theorems is that of Ulm [lo],One of the most notable classificationwhich classifies countable reduced abelianprimary groups in terms of a set of numerical invariants.This classi-fication using the same invariants has been extended to a much largerclass of abelian p-groups, the class of totally projective groups, inthe work of Nunke [g], Hill [5] (see Griffith [3]), and Crawley andHales [l].Hill introduced the concept of "nice subgroup" and established thecharacterization of a totally projective group as a reduced p-group thatcontains "enough" nice subgroups.Moreover, Hill was able to show thatthe class of totally projective groups is the largest reasonable classof reduced primary abelian groups that can be classified by their Ulminvariants.~f G is an abelian p-group (p prime) and n is a non-negative integer,then G [ ~ ] denotes the subgroup of G consisting of all elements havingorder less than or equal to pn; ~[p] is called the socle of G. In thisnpaper a necessary and sufficient condition is found for G[p ] to be anice subgroup of the reduced abelian p-group G (see Theorem 1). Thiscondition, together with the result of Hill [4], leads to the character-ization of a reduced bounded p-group as a reduced p-group in which allsubgroups are nice (see Theorem 4).Pfie^Mn.inCm.&'iAll groups in this paper are assumed to be additively written re-duced abelian p-groups.order equal to a power of the prime p.)- ~(A p-group is one in which all elements haveA group G is a divisible groupl~he author wishes to acknowledge the help of K. D. Wallace in writingthis paper.

J f for each x E G and each positive integer n, there exists y E G wchthat x = ny.ble subgroups.A reduced group is me which contains no (nonzero) divisi-(Clearly the element 0 is divisible by every positiveinteger n, since 0 = no.) Now let G be a p-group for the prime p.each ordinal a, we define inductively a subgroup paG as follows:pG = {px 1 x E GI; pY+G = p(pYG); if y is a limit ordinal pYG = n pG.KYWe thus have a descending chain of subgroupsG =pG 2 zPaG =p% 2 .Note in particular if n is a positive integer and x F pnG\pnlG then xis divisible by pn in G but x 1s not divisible by pn+l in G.If x isdivisible by pn for every positive integer n, then x E puff where u isthe first infinite ordinal and x is said to have infinite height in G.In general, the element x is said to have p-height a in G, and we seth (x) = a, if x E pa~\spa+l~. If x E paG for each ordinal a then h (x) = =GGwhere - > a for each ordinal a. Since 0 e paG for each ordinal a,hG(d =The following fundamental properties are easily established.G be a p-grcup with x,y E G.If h(x) # hG(y) then hG(x + y) is thesmaller of the two heights. If h (x) = hG(y) then hG(x + 9) S hG(x).GA homomorphism cannot decrease height; that is, if f is a homomorphisma+1from G into G', then hG(x) 5 hG,(f(x)). If paG = p G, then pBG = paGfor 6 > a and paG is a divisible group.there must be an ordinal y such that pYG = { 01.If G is a reduced p-group,y is called the length of G and is denoted by \(G) = y.ForLetThe least such ordinalA bounded groupis a torsion group for which there exists a finite upper bound to theset of orders of all elements. That is, there exists a positive integern such that nx = 0 for every x in the group.Alternatively, G is abounded group if there exists a positive integer n such that nG = 0.A subgroup A of the group G Is a niee subgroup of G if and only iffor all a. We note that the subgroup A of G is nice if and only if foreach x E G there exists some a E A such that hGIA(x + A) = hG(x + a).The following characterization, given by H i l l [s], will be utilized:Theohan A:The subgroup A of a p-group G is a nice subgroup of Gif and only if each coset x + A contains an element x + a that hasp-height in G which is maximal among the elements of the coset x + A.An immediate consequence of the above theorem is the fact that anyfinite subgroup is nice.Theorem 1:Let G be a reduced abelian p-group and n a positive in-teger. Then G[pn] is nice if and only if \(GI Â u.Proof:Suppose n is an arbitrary but fixed positive integer, G[yn]is a nice subgroup of G, and that A(G) > ID. Consider the descendingchain of subgroupsG 3 pG 2.. .3 pmff 3.. .3 pWG 3 pu+'$ 3.. . .u+1Since A(i7) > u, puff # { 01 and there exists x e p^G - p G, x # 3. Notethe height of x is u. Thus for the positive integer n there exists annelement y in G such that x = p y. Moreover, for each positive integer mx E pn+m~ = pn(pmG)and hence there exists xnE pmG Such that X = p xm. NOWmnp ( x - y) = pnxm - pny = x - x = 0so x - y E G[pn] and thusmx + ~[p"] = y + G[pn] .mSince G[pn] is nice, the coset y + G[pn] contains an element y + a ofnmaximal height. Hence hG(y + a) S hG(y + g) for each g E G[p 1. Sincex E y + G[pn], h (y + a) > h (x ) 2 m for each positive integer m. Thusm G G mhG(y + a) S u. Now observe that since a e G[pn], pa = 0 and thereforepny = pnY + pna. T~USn= h (x) = h (pny) = h (pny + pna) = h(p (y +a)) 2 ID + n,G G Gwhich is absurd. (Indeed, h (px) 2 h (x) + n holds in general.) Thus ifGGG[pn] is nice then \(G) < u for each positive integer n. If \(GI < u,then the set consisting of all ordinals which serve as heights of ele-ments of G is finite and consequently any set of elements of G must con-tain an element of maximal height.non-negative integer n.In particular, G[pn] i s nice for eachNow suppose \

a7r ~[p"], there exists an element a E G[pn] such that hG(x + a) < hG(x + z).Consequen:ly for each positive integer N there exists a e G[pn] such thath(x + a ) > N. If a E G[pn] then pn(x + a) = p x and hG(pnx) =hG(pn(x + a)) 2 hG(x + a).Thus hG(6'x) 7 hG(x + a) for each a E G[pE]and consequently hG(pnx) > N for each positive integer N.pnx f 0 and hG(pnx) > ID,Therefrrecontradictory to the assumption that A(G) = ID.Thus if A(G) 5 ID, G[pn] is a nice subgroup of the reduced p-group G.A topology may be introduced on a reduced p-group G by letting thesubgroups {pn~},z

mit follows that Z(p ) is an epimorphic imape of G.Theohem 4: Let G be a reduced p-group. Then G is a bounded groupif and only if each subgroup of G is a nice subgroup of G.Proof: If the group G is bounded then A(G) < lsi. Hence any set ofelements has an element of maximal height which implies that every sub-group is nice. Assume that every subgroup of G is nice. This implies,by Theorem 2' with C the class of all subgroups of ~[p], that the groupG is a direct sum of cyclic groups.Then G is either bounded or unbounded.If G is bounded we are finished. Assume G is unbounded. Then by Lemmasm2 and 3 there exists an epimorphism from G onto Z(p 1, say Y : G ->Ã z(pm).Note Ker Y # { 0) and is a divisible group (since z(pm)).Ker YKer YLet x i Ker Y; then - = h (x + Ker Y) > hG(x + k) for any k E Ker Y.-Ker YThus Ker Y is not a nice subgroup of G which is a contradiction of ourhypothesis.is a bounded group.Hence if each subgroup of G is a nice subgroup of G then GREFERENCES1. Crawley, P. and Hales, A. W., "The Structure of Abelian p-groupsGiven by Certain Presentation," J. of Alg. 12 (19641, pp. 10-23.2. Fuchs, L., Abelian Groups, Publishing House of the Hungarian Academyof Sciences, Budapest, 1958.3. Griffith, P., Infinite Abelian Group Theory, University of ChicagoPress, 1970.4. Hill, P., "Sufficient Conditions for a Group to be a Direct Sum ofCyclic Groups," Rocky Mt. J., 1, No. 2, (1971), Pp. 345-351.5. Hill, P., "On the Classification of Abelian Groups," to appear.6. Kaplansky, I., Infinite Abelian Groups, University of MichiganPress, Ann Arbor, 1954.7. Kolettis, G., "Direct Sums of Countable Groups," Duke Math. J.,27 (19601, pp. 111-125.8.-Megibben, C.. "On Mixed Groups of Torsion-free Rank One," IllinoisMath. J. , 11 (1967), pp. 134-144.9. Nunke, R., "Homology and Direct Sums of Countable Groups," Math. Z.,101 (1967), pp. 182-212.10.. Ulm, H., "Zur Theorie der Abzahlbar-unendliched Albelschen Gruppen,"Math. Ann., 107 (1933).11. Wallace, K., "On Mixed Groups of Torsion-free Rank One with TotallyProjective Primary Components," J. of Alg. 17 (1971), pp. 482-488.USE OF MATRICES IN THE FOUR COLOR PROBLEMby Ualuil tU.notiPoQte.cl~~c 11~tULu-te. of, NW YohkThe four color problem has fascinated mathematicians and laymen forhe pasr 100 years.Notwithsranding rhe massive research in this areaand rhe rampanr growrh of graph ~heory, of which map coloring is a sub-field, he four color problem remains unsolved to the present day. Thereader may wish to consult Marshall [lland May [5] for a historical ac-count of the subject.The four color problem may be stated with the aid of the followingtwo definitions:Oef,.b%&ion 1: A finite collection h! = {R.} of closed subsets ofi.E^ having the propertiesand1) each R. has positive area,2) Ri n intR = $ foralli and j, i # j,33) each R. is a homeomorphic image of the unit circle,is called a map; the Rijqs are called regions.More refined definitions can be given using the concept of planargraphs; see Ore [3], Tutte's definition in Harary [4].Uef,i.fion 2: A wit-coloring of a map is a function assigningto each region R. e a color a suah that, for all j # in, if1- o io'3R n 3R is a set of positive measure then c # a1-0 3 io j'The central question is: Given a map M uhat is the minimum number, x,,,of colors which is required to well-color the map?Obviously, if themap has n regions, xM 5 n; however we can do much better than this.fact, as seen in Ore [6], for any map h!X\,( 5 .We may now ask: Is it possible that four colors are sufficient to well-color an arbitrary map h!?An obviously related question is: Can we con-struct a counterexample which cannot be well-colored with four colors?In

Amazingly, neither of these apparently simple questions has yet been an-'zwered, and this is what constitutes the so called Four Color Problem.In this paper we definitely do not solve the four color problem,bur simply offer a new approach TO rhe problem.region-region marrix H = [a. .] as defined as follows:1-3acj = 1 i f 3R. n 3R. is a set of positive measure,= 3aij = 0 if aR. n 3R. is a set of zero measure,T-3To every map assign aFor a map with n regions the matrix H will be n x n and, as indicatedlater, it contains certain information concerning the map.(This notionis similar to that involved in a general graph when one constructs theso called adjacency matrix.)that :It follows directly from the definition1) The marrix associated virYi any map is symmetric.ii) The diagonal entries must be ones.iii) A column cannot consist entirely of zeroes.Iv)In any row there arc at least two non-zero entries.Property (i) implies that we need to study only the upper triangularmatrix.An example follows.The upper triangle isThe matrix can be read as follows:1 1 1 0 11111I 11111.Region 1 touches regions 2, 3, 5 butnot 4 (read row 1 across); region 3 touches regions 4, 5 (read row 3across) and regions 1, 2 (read column 3).Since the numbering of the regions in the collection f.1 is arbitrary,we can use different numberings. This induces different matrices. Thereare in fact n! "different" matrices for a given map.If Ho is any parti-cular matrix, the remaining n! - 1 matr:ces are called derivatives of Ho;symbol:anr: - any derivative of Hn.We are now ready to define an equivalence relation between maps.The following is easily seen to be an equivalence.Ve/,.in^tion 3: hfl "i. h19 if the set of derivatives of a matrix ofMI coincides with the set of derivatives of a matrix Hi of /.la.Obviously a necessary condition for two maps to be equivalent is'that they have the same number of regions. On the other hand, two mapsMI and hf-; being homeomorphic is sufficient to insure MI hf;. For exampie,the two maps given in Fig. 2 are equivalent and consequently areconsidered to be indistinguishable.FIGURE 1In the map illustrated in Fig. 1, we havey(3Rl n 3Rl) > 0, all = 1,-.y(aRl n 3R2) > O., a;? = 1,u(aRl n > 0, a13 = 1,- -- v(aRl n 3R4) = 0, a14 = 0, and so-forth.Hence the matrix associated with this map isFIGURE 2The converse, however, is not true, as Fig. 3 shows. These two naps areequivalent (as can be seen by writing out the 4! = 24 matrices asscciatedwith each map) but are not homeomorphic, since in the sec0r.t 7-223Rz n 3R4 # 4.

FIGURE 3Given two equivalent maps, if we well-color one map and assign anumbering to the second map such that the matrix thus generated is equalto the matrix used to color the first map (we show below how this isdone), then we automatically have well-colored the second map.Hence thechromatic characteristics of maps belonging to the same equivalence classare the same.For this reason we have only to study the equivalenceclasses of maps, which is equivalent to the study of the associated ma-trices, or in particular, any one of the associated matrices.The equivalence relation introduced above reduces the number ofmaps in E~ having n regions to a finite number of equivalence classes.The number of equivalence classes for maps with n regions is bounded byl/2n(n-1)(use properties (ii), (iii), (iv) of the region-region matrix).The function f associating the region-region matrix to a map is a func-tion into the set of symmetric matrices.matrix111111111111111I l l111It will be seen later that thehas no pre-image; for if it has a pre-image, it would require 7 colorsto well-color it, but, as we have indicated, a:,,, 5 5 for any map M.Now we introduce a technique to well-color a map using the associ-ated matrix without any reference to the actual map itself. We have alreadystated that we need investigate onlv the upper triangular matrix--one canconstruct the triangular matrix by determining, for a particular regioni, the regions with assigned integer greater or equal to i, which havea common border with region i. The well-coloring procedure is as follows:Suppose that we make the first region of color A.first row by A.We then "multiply" theFor the preceding matrix map of Fig. 1 we would obtainA A A O A1111I l lNow investigate the second (ith) region.ond (ith) region of the same color A11J . .We are allowed to make the sec-as the first region (the same coloras the (i - l)st, (i- 2)nd-0-2nd, 1st regions) ONLY if the second (ith)column does not already contain color A (color of (i - l)st, (i - 2)nd- --etc.) We see we cannot use color Aond column already has Aand hence "multiply" the row by B.for the second region since the sec-in It; we must use a different color, say B,We getA A A O AB B B BI l l11Continuing this process we finally obtainA A A O AB B B Bc c cNote that in row four we can use color Aagain, since that column doesnot already contain A. This means that region 1 will be made color A;2, B; 3, C; 4, A; 5,WID. If we now go back to the actual map, we see thatit is well-colored.FIGURE 4

Oe;(i&on 4: The cohati.on of a matrix is the assignment of alabel L(a..) to every non-zero element a of the matrix.t'jMVedin&ion 5: An n x n matrix iM with a coloration is iwrrnaz if1) for all i, 1 5i 5 n, L(a. } = L(a..) for all non-zero elementsIt It"it '2) for all j, 1 5 3 < n, L(a . .) # Ua ) for all non-zero elements33 tiat., t< 3.Theohem 1 : Let H be some matrix associated with a map M. Then f.1is well-colored if and only if H is normal.Proof: Let the matrix be normal; the color on the diagonal is dif-ferent from any color in that column.Consider a... The fact that ail,33a3, aj, ..., ajj are, say, 1's means that the (jth) region touches theIst, 2nd,---, (j - 1)st region; but by hypothesis the color of the jthregion is different from that of the lst, 2nd,--., (j - l)st region, con-tiguous TO it. This being true for all columns implies rha~ rhe map iswell-colored.Conversely, let the map be well-colored.Then any region j is sur-rounded by regions with colors different from that of the region j. Givean arbitrary numbering to the map and also write down a letter symbol forthe color in the appropriate region.Now construct the associated matrix,but instead of 0's and l's, put 0's and the letter of the color.the (uncolored) matrix is of the form1 1 1 0-0.1 11 1 1 - e . 1 11 1--.1 1Say thatBy hypothesis all regions around r'egion 3, say J' = 3, have differ-ent colors.This means that the entries in column j, have differentcolors (by construction of the matrix).form1 1 A 0---1 11 B 1."1 1C 1---11and hence by the multiplication principleSo the matrix will be of theA A A 0"'A AB B B"-Bc C---Cand it is normal (the argument above being true for all columns).A question of considerable importance is if the map is well-coloredwith the minimum possible number of colors.BcObviously, normality doesnot imply that the map is well-colored with the minimum number of colors;the converse Is however true.In order to well-color with minimum numberof colors, every time we get to a new row we must investigate to see ifwe can use again colors already used..Sometimes we will have a choiceof two colors (already used) leading to a different number of total colorsused. In any case the matrix will be normal. Example:A A O O A A A 0 0 AB O B S B O B BA A A B B Bc C A APc(4 colors needed) (3 colors needed)This does not present a problem, however, since In application of thisprocedure to the four color problem we suppose there exists a matrix withn + 1 columns, representing the smallest map h!in E? requiring five colors.If we can fuse together two regions in such a way that the resulting n x nmatrix (map) still requires 5 colors, we achieve a contradiction sinceII was the smallest such map; hence no such map could exist.we already know that n has to exceed 40.Of course,See Ore and Stemple [?I. Tothat end we present (without proof) the following theorem.Theohem 2: Let ?.I be a map, /f its matrix. Let R , R t"S two con-tiguous regions (Le. a = 1) and assume j > a. Let A!' be a map iden-CVrical with 1.1 except that the regions R and R are combined. Let H' be a3a matrix obtained as follows:1) "Add" the corresponding entries of row j In H to row a in H asfollows: 1 + 0 = 0 + 1 = 1, 0 + 0 = 0, 1 + 1 = 1.2) "Fold" the J'th column at entry a+ Then:a) "Add" (as above) the corresponding elements of the unfoldedpart to the entries of the a columns

) "Add" (as above) the corresponding elements of e he foldedand finallypart to the a row.11 11111111 1 0111It should be noted that even though one row and one column disappeared,this fusion "swallows up" zeroes in the remaining rows and columns.The presence of more one's tends to increase the number of colorsneeded to normalize the matrix. Hence the proof of the following con-3) All other rows and columns of H remain unchanged.jecture, as indicated earlier, would solve the four color problem.Then there exists a numbering of the regions of hf' such that the matrixof M' is exactly H'. Conje.&tu~e.: Given an n + 1 matrix which requires a minimum of 5E X : Fuse region 2 and region 4 in map M below:colors to be normalized, then there exists one fusion of regions suchthat the resulting n n matrix requires 5 colors to be normalized.The above, and this we believe is the merit of this paper, is voidof any geometric concept and constraint: Given a matrix which requires5 letters to be normalized, fuse two rows (in the sense of the precedingtheorem -- without reference to geometry) such that the resulting matrixstill requires 5 letters. The efforts of the author toward settling theabove conjecture have not yet produced satisfactory results.H . = 1 0 1 1 1 11 om1 01 1 1 0I l l111FIGURE 5H 1 = l l l l l11111 1 0The matrix H' is obtained from H by the following steps:REFERENCESHarary, F. (editor), Graph Theory and Theoretical Physics, AcademicPress, New York, 1967.Harary, F., Graph Theory, Addison Wesley, Reading, Mass., 1969.Liu, C. L., Introduction to Combinatorial Mathematics, McGraw-Hill,New York, 1968.Marshall, C. W., Applied Graph Theory, Wiley-Interscience, New York,1971.May, K. 0. , "The Origin of the Four Color Conjecture ," Iris, 56,(19651, pp. 346-348.Ore, 0. , The Four Color Problem, Academic Press, New York, 1967.Ore, 0. and Stemple, Joel, "Numerical Methods in the Four Color Problem,"Journal of Combinatorial Theory, 8, (1970), pp. 65-78.

EULER'S CONSTANT AND EULER'S eThe sequence {adefined byna = x $-:nni=1can be shown to converge by demonstrating that the zequence is (1) de-creasing and (2) bounded below by 0. The value of this limit is oftenreferred to as EuZer's constant. Another constant associated with Euleris e. Euler introduced e to represent the base of the natural logarithm.One of the familiar properties of e is the following inequality:1 i 1 i+1(A) (1 + 7) < e < (1 + 7) , i = 1, 2, 3, .--.The purpose of this note is to show that by using (A), one can show thatthe above sequence (a 1 is convergent.nSince by (A)1 n+1e

PARTIAL DIFFERENTIATION ON A METRIC SPACE 1by RoAeann t:\o^L&UoSvton HaU. U~uvvui.UyThe "metric differentiability" of functions between abstract metricspaces is a relatively new and undeveloped field of mathematics.metric derivative of a function with real domain was first defined in1935 by W. A. Wilson (See [2]). In 1971, E. Braude arrived at the samedefinition after having researched the topic independently.TheUnlike hispredecessor, Braude sought to develop his concepts of metric differenti-ability for functions with abstract metric domains,In this paper weexplore the idea of partial differentiation for functions defined on thecartesian product of two metric spaces.PaJvtioJi cii.f.f.e~~e.ivU.a^i.ona mvUu.c ApaceLet f be a function from a metric space (X,s) into a metric space(Y,t) and let bX. f is said to be metrically differentiable at b ifb is not discrete and if a real number f(b) exists with the propertythat for every E > 0, a positive number 5 can be found such that ifs(xt, b) < 6, s(xV, b) < 6 and x' # x", thenThe assertion "f: X -* Y is metrically differentiable" means that f ismetrically differentiable at every point of X. (See [2].)Proceeding almost directly from the definition of the metric deriv-ative is the definition of the metric partial derivative.Let f be afunction from a metric space (XxY,p) into a metric space (Z,p ') and let(x,~) be an element of XxY for which every open ball containing (x,y)contains an element (xl,y) with x' # x; f has a metric partial deriva-t ewith respect to x at (x,y) if a rea-1 number /(x,y)exists with theproperty that for every n > 0, a positive number 5 can be found such thatif-p(xl, x) < 6, p(xr', x) < 6 and xt # xu, then'one of several papers produced as a result of a research project fundedby the Research Corporation under the Cottrell College Science Program,grant number c-205/308, directed by Professor Em 3. Braude, Seton HallUniversity.-.I^ pf(f(x', Y), W,IY) - y )

close to (xY y) it can be said that the quantityg(f(xfy g)* f(xffy 9))g((xty y)Â (xffy 9))gets arbitrarily close to f(xy y) for xf and xff sufficiently close tox. Thus(3) The following is true:(a) g((xf, yfIy (xf, y= h(CÂ c) = DYThe restrictions placed on g and h by the preceding lemma allowfor a more general statement of Theorem 1.Thtotm 2:Let (X, p) be a metric space, and let h be a continu-ous function of Rt x R' into R+ satisfying the conditions on h of Lemma lyas well as the conditions h(a, 0) = a, h(Oy b) = by and ch(a, b) =h(ca, cb) for every c, a, and b in R'. Let g be the metric on XxX de-fined in Lemma 1 by g((xf, yf), (xff, yff) = h(p(xf, xff), p(yf, yff). Thenfor every metrically differentiable function f: (XxX, g) + (XxX, g) givenby f(xy y) = (V(x, y), V(x, y)) for which

= lim hp(U(xf, y), U(xffy))Ip(V(xl, Y), v(xrrY Y ))xr+ x p(xfy xrrl p(xf, xr9= h(max(b t cy e t f) = h(b t c, e t f).(I+) a 2 b and c 5 d imply max(ay c) 2 max(by dl. Nowyc(max(a, b) ) = max(ca, cb) since "d 2 kl' and "cd > ck" areequivalent for c 2 0.h(n, b) = b are clear.The equations h(a, 0) = a andApplying Theorem 2 to the function h of Remark 2 gives the follow-ing result.Coko.&by:Let X be a metric space with meeic p and let g be themetric on XxX given by g((xr, yr), (xrry yrr)) = max@(xl, xrr), p(yr, yrr)l.Furthermore, let f:(XxX, g) + (XxX, g) with f(x, y) = (U(x, yIy V(xy y))where U,V : XxX -+ X and

A SUGGESTION FOR ENJOYABLE READINGProfessor 20bert k!. Prielipp of the University of k!isconsin, Oshkosh,highly recommends three relatively little knobm books to Journal readers.Fantasia Mathematics, assembled and edited by Clifton Fadiman, Simonand Schuster, Inc., New York, 1958 (Paperback). Noteworthy selections inthis collection of short stories and poems are:(a) Young Archimedes, by Aldous Huxley -- the difficulties faced bya young Ttalian peasant boy who seems to be a mathematical genius.(b) The Devil and Simon Flagg, by Arthur Porges -- a mathematicalversion of Stephen Vincent Benet's The Devil and Daniel Vebster.(c) --And He Built a Crooked House, by Robert A. Heinlein -- unusualphenomena in a house having dimension higher than three.(d) A Subuay Named Moebius, by A, J. Deutsch -- strange things happento the Boston subway when the Boylston shuttle is installed.The Mathematical Magpie, assembled and edited by Clifton Fadiman,Simon and Schuster, Inc., New York, 1962 (Paperback). Choice selectionsin this anthology are:(a) The Lau, by Robert 14. Coates -- unlikely events become the rule.(b) The Appendix and the Spectacles, by Miles J. Brown, i.1.U. -- amathematician uses the fourth dimension to get even with a bank president.(cl The Nine Billion Names of God, by Arthur C. Clarke -- Tibetonmonks try to use a computer to set down all the possible names of God.(d) Milo and the Mathemagician, by Norton Zuster -- the adventures ofa little boy and a dog (who ticks) in Di.gitop01is.'-- +-Whom the Gods Love (The Story of Evariste Galois), by Leopold Infeld,McGrZw-Hill Book Company, Inc., New York, 1948 -- the moving story of ayoung mathematical genius who is caught up in the struggle for a new Frenchrepublic in the days after Napoleon's defeat during the reign of the Bourbons,His father commits suicide, his mathematical accomplishments gounrecognized during his lifetime, and at age twenty he is killed in a duelresulting from a love affair plotted by his political enemies.WELCOME TO NEW CHAPTERSThe Journal extends its welcome to the following new chapters ofPi Mu Epsilon which were recently installed:ALABAMA EPSILON at Tuskegee Institute, installed by Houston T.Karnes, Council President.CALIFORNIA IOTA at the University of Southern California, installedby J.C. Eaves, past Council President.GEORGIA GD4MA at Armstrong State Collegeby Houston T. Karnes.installed April 2 197QILLINOIS EPSILON at Northern Illinois University, installed by J.Sutherland Frame, past Council President.ILLINOIS ZETA at Southern Illinois University, installed June ly1973 by Houston T. Karnes.Eaves.T. Karnes.KENTUCKY BETA at Western Kentucky University, installed by J.C.LOUISIANA MPPA at Louisiana Tech, installed May 9, 1973 by HoustonMICHIGAN DELTA at Hope College, installed by J. Sutherland Frame.NEBRASK.4 BETA at Creighton University, installed April 25, 1973 byR-V. Andreey Council Secretary-Treasurer.PENNSYLVANIA MU at the University of Scranton, installed by EileenL. Poiani, Councilor.SOUTH CAROLINA BETA at Clemson University, installed March 27, 1973by Houston T. Karnes.

LOCAL CHAPTER AWARDS WINNERSThe Journal inadvertently omitted the 3LIFSRNIA ETA (Yniversity ofSanta Clara) report of their annual awards in the last issue.frr Excellence for 1973 was presented toK&tn A. !8foncta,and the George W.The AwardEvans 11 lrlemo~ial Prize for highest placing in theWilliam Lowell Putnam Mathematical Competition was received byPad N. lbcquaKdUttn hi. Vdy.The winner of the Freshman Mathematics Prize for 1973 wayJme~ L. Hahna.A FRIENDLY REMINDERWe are no longer printing the long initiate lists in the Journal,but we do print the names of those who have distinguished them-selves in mathematics in their local chapters.PLEASE SEND US THENAMES OF YUL'R AWARDS WINNERS and let your exceptional chapter mem-bers (or local students) will receive the recognition they deserve.In connection with this7 remember that the National Office cansupply you wiyh impressive award certificates.Write to:R. V. AndreeSecretary-Treasurer 7 Pi Mu Epsilon601 Elm Avenue, Room 423The University of OklahomaNorman, Oklahoma 73069NO ANNUAL MEETING THIS SUMMERBecause the annual summer meeting of the Mathematical Associationof America wL11 not be held this year, Pj Yu Epsilon also will not haveits meeting.The Council urges as many local chapters as possiLle tomake plans to send delegates to regional meetings of the llAA in theirarea instead.Many of these regionals include sessions for undergraduatepapers, so be aware of this opportunl+y for your members.PROBLEM DEPARTMENTThis department welc~~mes pr~blems believed to be new and, as a rule,demanding no greater ab:lity in problem solving than that of the averagemember of the Fraternity. Occasionally we shall publish problems thatshould challenge the ability cf the advanced undergraduate OP candidatefcr the 1.laster's Degree.methods of solution are also acceptable.q?d problems displaying novel and elegantProposals should be accompaniedby solutions, if mailable, and b3 an2 information that will assist theeditor.Solutions should be submitted on separate sheets containing thenme and address of the solver and should be mailed before the end ofNovember 1974.Address all commun

which is greater:- 12 tan-'(fi - 1) or 3 tan-'(l/~) + tan (5'99) ?31 7. Phopobed by tke E&oh 06 tke PhobLem Depa,ttinent.A rectangle ADEB is constructed externally on the hypctenuse AB ofa right triangle ABC (Fig. 1). The lines CD and CE inter~ect the lineAB in the points F and G respectively. a) If DE = AD^, show thatA G ~ + F B ~ = AB~. b) If AD = DE, show that F G ~ = AF GB.320. Paopv~cd by H. S. !.I. Cvxcteh, Tctoi~tv, Ca~lada.Prove that the projectivity ABCX BCD (for 4 collinear points) isof the period 4 if and only if H(AC, BD).321 . Pavpobcd by Nob~nv King, RdcLgh, Nv~th Cc~~fi~ta. (Dedicatedto .t/~e tnctnvag 06 Lco Lfvbm)According to bierten's Theorem.where y denotes Euler's constant, (9.57721 ... ) an^ where the product onthe left is taken over all primes not exceeding n, (See Hardy and Wight,The Theory of Numbers, p, 351 01- Trygve Nagellis Introduction to hnberTheory, p. 298).Can you estimateYoak,322. Paopobed bg Jack Gm~unke.L!, Fokut Hi.L& High School, NCWIt is known that the ratio of the perimeter of a triangle to thesum of its altitudes is gpeater than or equal to 2/&.,Vathematical i.IonthZy, $roblem E lk27, 1961, pp. 296-297).(See AmericanProve thestronger inequality for the internal angle bisectors t t and t :a' bFIGURE 131 8. Paopob ed by R. Ro biiaon Rome, Sacmne~do, Cafi~vaiua.Two equal cylindrical tanks, Tank A abcve Tank B, have equal ori-fices in their fl:>ors, capable of discharging water at the rate of 13figallons per minute, where h is the depth of water in feet.A t 10:2,, a.m.Tank B is empty and water 1s 19 feet deep in Tank A, as discharge begins.At noon Tank A is Just emptied, What was the maximum depth in Tank B,and when ?How deep is the water in Tank B at noon, and when will itbe srqty?-319. Paopobed by Pao6e~boa 1.1. S. LongueA-Uiggi~a, Catnb~dge,England.Let A', B', C ibe the images of an arbitrary pcint in the ?idesBC, CA, AB of a triangle ABC. Prove that the 4 circles AB':', BC1t!',CA'B' , ABC are all concurrent.equality holding if and only if the triangle is equilateral.323. Paopvbed by David L. Sdhmnan, Lob Angeh , CaLL6omua.Call plane curves such as the circle of radius 2, the square ofside 4? or the 6 x 3 rectangle in Fig, 2 bornetrfc if their perimeterts numerically equal to the area they enclose. \!hat is the maximum areathat can be enclosed by an isometric curve?FIGURE 2

"Whconhin.324. Ptopohed by R. S. Lut\m, U~uwmLtq 06 [ U ~ C O I I . ~ ~ IJanc~wLUe,I ,Evaluate325. Ptopohed by Cl~utLu (0. Th,igg, Salt Viego, Cdi6ot1ua.Show that there is only one third-order magic square xith positiveprime elements and a magic constant of 267.Solutions292. [Spring 19731 Ptopobed by Jack Gut6u1tk& Fohuf H ~ Y A HighSchool, Fhlung, Nu Yotk.If perpendiculars are ccnstruc-ed at the points of tanFen(:y cf theincircle gf a triangle and extended outward to equal length:,then theioins of the% endpoints form a triangle perspective with the given tri-angle.SoLCLon by Chyfon W. Dodge, U n i w ~ d 06 y !.faine at Oho110.FIGURE 3the incircle touch sides BC, CA, AB of triangle ABC at pointsX, Y, Z, (see Fig. 3) and let the other ends of the constructed equalperpendiculars be P, Q, R, respectively.It is well known that pointsA, B, C and X, Y, Z are in perspective at the Gergonne point for trian-gle ABC. But also triangles XYZ and PQR are in perspective at point I,the incenter for triangle-ABC.It is also known that perspective istransitive; that since triangles ABC and XYZ are in perspective and tri-angles XYZ and PQR are in perspective, then triangles ABC and PQR arein perspective .also.A.00 holwed by ZAZOU KATZ, Bewdy H i L h , Cdi6ohnia, and fhePtopoh WL.Ptoblm EcLLtotlh NoZeTWO recent texts describing the Gergonne point and other notablepoints of the triangle are:Rinehart and Ninston, Inc., 1969; (2) Clayton W.(1) David C. Kay, Cozzege Geometry, Holt,Dodge, EucZidean Geome-try and Transformations, Addison-Wesley Publishing Company, Inc., 1972.293. [Spring 19731 Phopohed by Lau Kot~mLi, !.10hgafl s m e College,W n o t e , hfmy&znd.Prove that N = 531° + is divisible by 78.1. SoLutio~t by R. Robimon Roiue, Saucamento, Cdi6otlua.Since 53 * -l(mod 6) and 103 : l(mod 6), we have 53lU3 E (-1)lo3(mod 6): -l(mo6 6) and 1 l(mod 6).Alsc, since 53 : l(mod 13) and 103 : -l(mod 131, we have 531° !l(mod 13) and 1 0 3 5 ~ -l(mod ~ 13).Hence N is divisible by both 6 and 13, which are relatively prime.Therefore fi is divisible by 6 . 13 = 7B.S.i.m&RAYMOND E.h o W o ~ u wme odimed by MERYL J. ALTABET, &onx, N. Y.;ANDERSON? !.lo~tta~ta State UniwmiAy, Bozana~~, !.lo&na; DONALDCASCI, Rltodt Zh.f!.und College, Phowidence, Rl~ode Zhhnd; RICHARD A. GIBBS,Foa Laud College, Quha~tgo, Colomdo; DONNELLY JOHNSON, hlajoh, USAF,A h Fohce ZnhLLtuLe 06 Tech~~oLogy, Fahbot11, Oluo; BRUCE .LOVETT yRuLg~College N u Bhu~t~wick, N. J.; T. E. MOORE, LVLL~~~LU&VL SXuAe College,Wdg watm, !.lab.; THERESA PRATT, N. Eafon, blab. ; PAOLO RANALDI ,Aho~, Oluo; MICHAEL SCHWARZSCHILD , PoLyfechnic 71%+fi't&te o 6 BhookLyn.11. SoLCLon by C l M u (0. Th,igg, Sun Viego, Cdi6otnia.Examining the congruences of 531° and 1 0 3 for ~ ~ the moduli 3, I+,

7: 13 and 25, it is seen that N is divisible not only by 78 = 2(3)(13)but also by any combination of the factors 2', 3, 5', 7, and 13.S o U o n ~ tecognizing tL.4 e&eaion 06 the ptopobed ptobLan WJL~066e~d by CLAYTON W. DODGE, U~uumLty 06 !.laine and by DAVID L. SILVER-MANy Lob Angdu, C&6otni.a.111. SoUon by Funk. h!U5bhn0 and Almk. Yank.ouich, Juniou at VtudUniuuudy, Phdkddphia, Pen~uyLuanh.Expansion of (78 - 25)lL3 and (78 + 25)53 by the binomial theoremshows that each term contains a factor of 78 txcept (-25)1''3 and w3.Therefore, if 2sS3 - 25173 is divisible by 78 then so is N. Rewrite2553 - ;51:,3 as -(25)53(255c - I), which equals -(2~)~~(625~~ - 1). Itis known that (62sZ5 - 1) is divisible by (L25 - 1) = ? x 78.Therefore78 divides 2sS3 - 25lo3 and also 531° + 103~~.0 t h 6oUo1u ~ ~ inuoLving the binomid tl~eotuti tuem o66ehed byHYMAN CHANSKY, UvuumLty 06 AImyLand; STANN CHONOFSKY, Wotca,ta PoLytechnic1nAUute, Wotcatm, blab.; JACK GIAMMERSE, LOLLLLLUVIU S m eUniuwLty, LWon Rouge, Louhiana; PETER A. LINDSTROMy Genuee Co~mnuni-ty CoUege, LWauia, NU Yotk; T. PAUL TURIEL, SUNY at !3ing\mntob1, N ~ uYotk; CHARLES W. TRIGG, Sun Diego, C&dot~%iu; and the Phoj20.4U~.CommentGeneralizaticns of the problem were subm!ttedby Theodore J~n~~eis,Brooklyny New York and by Bob Prielipp, Univ*rsity of \lisconsin at Osh-kosh:If p = a m + 1 = a n - 1 and q = aqm - 1 = a,,n + 1, and p and q1 2are odd, it is found that pq * 8 is divisible by 2mn.294. [Spring 19731 Ptopobcd by CMu [U. Th,Lgg, Sun Diego, CULL-6otni.a.In Fig. 4 show that ABC3 is a square.FIGURE 4Comment by Chy.ton W.Dodge, UnLumiAy 06 !.lalne at 0to110,The solution of the analogous problem for nested equilateral tri-angles appeared in the November 1970 issue of the f.Jathematics 14aqaaine,Problem 754, pages 280-281.Comment by K, R, S. Sastry, Xakele, Ethio-pia, in the November 1971 issue indicates that the method of the solutionby 1-Iichael Goldberg holds for any regular polygon.Hence it holds fora square, proving problem 294.SoLuLio~u tume &~o 066aed by ZAZOU KATZ, Beudy H&, Cafi6ot1ua;ALFRED E. NEUMAN, ALph D&a FmtemLty; R. ROBINSON ROWE, Sammento,CaLitjot~ua; and .the Ptopob u.295. [Spring 19731 Ptopobed by /blmay S. K ~ J I I ~ ~ Fotd J I , Scie1tti6.L~Labotatoty , Uembotn, !.liclugan.Determine an equation of a regular dodecagon (the extended sidesare not to be included).SoUon by ,the Ptopob a.The equationNn=lrepresents a 4N-gon.lanx - b,?\ + [cnyn - dnl> = ABy symmetry we trylx - 11 + :x + 11 + lg - 11 + l yAfter sane elementary calculations,a = 5- 1 >, = 2(2+ 6) .+ 11 + ~(1x1 + Iy!) = A .296. [Spring 19731 Ptopobed by SOLOIIIOII (t!. GoLo~nb, U~uuemLty 06SOUZ~I~II Cal56ot1ua, Uejmdmeld 06 ELect~~Lcd E~tginedng.1) Combine 2, 5, and 6 to make four 2's.2) Combine 2, 5, axd 5 to make four 4's.3) Combine 2, 5, and 6 to make four 5's.4) Combine 2, 5, and 6 to make four 7's.5) Combine 1, 5, and 6 to make four 7's.SOLULLOII by ChmLeb (U.In general,6 + 2 - 5=3=(a+a+a)/a;Tfigg, SUII Diego, Cu& 6ot1ua.2(6 - 5) = 2 = a/a + a/a = 6 + 1 - 5;2 + 5 - 6 = 1 = (a/a)(a/a) = aa/aa = a/a + a - a = aa/aa = l(6 - 5);6-5-!2=O=a/a-a/a=a+a-a-u=aa-aa=aa-aa=6 - 5 = 1 ;

where a is any positive digit, including 2, 4, 5 and 7. (The qymb51 !xindicates sub-factorial x. For example: !l = 0, !2 = 1, and !=, = 44.)Attention is called to the solution of Problem E861, American MathematicalMonthly, January 1950, page 35, where Vern Hoggatt and Len b1r:sershow that every integer may be expressed by using p a's and a finitenumber of operator symbols (including log) used in high school texts;p > 3 and a > 1 are integersFor other representations where a has a specific value, space isconserved by first listing certain non-negative integers with their representationsby 2,

Using toothpicks and rearranzing:- Ncte:bined so as to equal the sum of the four like digits.PhobLem Eddoh'b CornmeldIn each of the five cases, the three unlike digits have been com-The multitude of solutions listed belowy in some cases duplicatingrepresentations already listed in trig^'^ solution, display the ingen:ousvariety accomplished by the following solvers: CLAYTON W. DODGE, UnLum.Ltqo# ;.Iaine a,t Oh0n0; JOAN INNES, ch&ighton UnLwui~.Lty, Omaha,Nebhuh ka; RICK JOHNSON, Bmougb cohpo~on, Whington, N. C. ; BRUCELOVETT, Rdgm CoUqe, Nuu B J L U I ~ ~ N. C ~ J.; , JIM METZy SphingdieLd,IL&no.Lh; T. E. MOORE y Wdge~uatm State CoUege, Eitidgui~2.m. !.la~ach u m ; BOB PRIELIPP, The UnLumLty 06 W.Lhconhin, Obhfzo~h; R. ROBIN-SON ROWE, Sawento, C&6oh1&; and the Phcpo~m.The potpoumi 06 ~o.fdLonh bubmitted by the above-mwd AOLLJW .Lh a6oUom :1) 2 ~ 5 + 6 = 2 - 2 ~ 2 * 2 = ~ ~ * ~ ~6/2+5=2+2+2+2256 = [(22)21~ ; 2(.5 6, = 2 + 2 + 2 +. : = 2A + 2:152161 = -1-6:/5] = 2 t 2 + 2 t Z ,where brackets Indicate the greatest integer function.297. [Spring 19731 Ptopo~ed by Rogm E. KueL, Kunha City,ili~bod.A traffic engineer is confronted with the prob:em of connectingtwo non-parallel straight roads by an 5-sha7ed curvs: formed by arcs zftwo equal tangent circles, one tangent to the *:rs? road dt a selectedpoint and the other touching the second road a-FIGURE 5a cave-. ?~int.1) &termhe the radius of the epal chle~ synthet~cally, zr2g-e#dmetrically w analyt3cally.2) If the f e e lends it~elf ta an Ew$idean c~nSructio3, bxMuld Hhe g6 about $t?PhobLem Eddoh'~ NoteBecause of the ehple#tyand dyffh~lty of this problm :he dead-line for consideration of solutions is hereby entende? TO !-io*.*e&r 30,1974. A synthetic solution would be particularly welcore.298. [Spring 19731 Phopobed by Pad Ehdzh, Budc~pe&,Jan hfq&ebL, UbuwmLty 06 CoLohadc, BO~~LYL, Cob.tadu.bove that11) lim ;( 6 t 3 6n-1g) lim ;(nl'lOg+ nn-SoluLLon by Vonn&yBuhe, Ohio.Let E > 0 be given.+ - -. + n&) = 1,J. JO~IIUOI~, hfajc~t, WZ,Using L'Fos~i~~l'sThus, there is a function hf( E) so ?hzznu i m ]L> 1 nl'k - I] < 5 i+k=2H

thenAlso observe thatso ifthen< log(1 +:) and nl'k < 1 +; .n= 2 log n 1/2l/Z 112 'Let us number the boxes in the tic-tac-toe array along each rowfrom left to right using 1 to 9, Thus the first column is 1-4-7. Ofthe 2 X's and 2 O's, it is clear that :1. lio mark (say X) can occupy box 5, since then a like mark isadjacent to it, so the opposite mark (0) must complete that row or columnor diagonal, The last 0 cannot now be placed so that a win cannot beforced in some way.2. No two like marks are symmetric to box 5, since then a like markin box 5 would win or observation 1 would be violated.3, The same mark (say X) cannot appear in two adjacent corners(say 1 and 3), since then 0 must occupy box 2 and some other box thatwould enable 0 to win if he plays next.4. The same mark (say X) cannot appear in two middle-of-an-edgesquares (such as 2 and B or 2 and 4) without violating observation 2(as in boxes 2 and 8) or, if we assume X's in boxes 2 and 4, then 0 mustoccupy the corner (box 1) between the X's and another box (either 6 or8) to prevent X from v~inning with box 5, But then 0 can win with box 9.5. The same mark (say X) cannot occupy two adjacent squares (say 1and 2), since then the opposite mark would have to complete that POW orcolumn ((2 in box 3) and another box, enabling 0 to win if he moves first,6. Thus each mark occupies one corner and one middle-of-an-edgebox (such as 1 and 6) that is opposite. There are four cases.If X lies in 1 and 6 and U in 2 and 9, then X in 4 forces a win forAbo boLved bq .tllc P&opobm,- -E E Ec-+-+-= E asrequired.3 3 3299, [Spring 19731 Pfiopobed bq Uciv.Ld L. S.Zvemtta~~, (Uu.t LobAngdeb, Ca&~o&~ua,4-- On the back of an envelope you see the results of an interruptedgame by two players whom you know to be tic-tac-toe experts,It isgenerally recognized that the expert nevgr puts himself into a puten-tially losing position and always wins if his opponent gives him theopportunity,The~e are 2 X's and 2 0's on the diagram.It is impos-sible to deduce whose move it is. Neglecting symmetry, what is theposition?x *x.If X lies in 1 and 6 and 0 in 2 and 7, then X in 5 forces a win forIf X is in boxes 1 and 6 and if 0 is in 4 and either 3 or 9, thenexpert play will produce a tie in all cases.Hence we have two distinct solutions to the problem* either of whichcould have been on the envelope:

ALAO solved by RICK JOHNSON, Wibnhjton, N. C..; ZAZOU KATZ,HeC.E.s, CatC.&o,z~ua; and the Pzoposc-z.300. [Spring 19731 Ptoposcd bg the. Piobicin EdLto,~.It can be shown with difficulty that if the opposite anglesskew quadrilateral are equal in pairs, the opposite sides are also eaualin pairs. (The reward of instant immortality is offered the solver whocan prove this without difficulty). If t:!o opposite sides of a ske:.!quadrilateral are equal and the other two unequal, is it possible tohave one pair of opposite angles equal?I. SoLtLon by Ctay.tun W. 'Dodge, l.litivt,'t~.Lty of !:la-i,ne. at (h.oiio.The answer is "yes." We make use of the "ambiguous" case SSA forcongruence of triangles. Let ABC be anv obtuse triangle with obtuseangle at B and with side AB greater than side BC (Fig. 6). V!ith centerB swing an arc of radius BC to cut side AC again at E. Then trianglesABC and ABE are not congruent but satisfy the SSA condition. Constructtriangle BCD (on side BC) congruent to triangle EBA. Now fold the paperalong line BC. Then the skew quadrilateral ACDB has angles A and Dequal, sides AB and CD equal, and sides AC and ED unequal.IT. SoL-Uon by Zazou Kats, Bwe~-Cy H,(¥£.Caf-L^o~i~C.a,Consider the annexed figure, in which chords CD and AB are parallel(Fig. 7). It is clear that angles DCB and DAB are equal and that CB = AD.if triangle CDB is hinges about DB and C is lifted from the plane of thefigure, we obtain a skew quadrilateral in vhich the angles at C and Aremain equal, the opposite sides BC and AD are eaual, and the oppositesides CD and AS are unequal.A£i ~otued biy ALFRED E. NEUMAN, (In A£p V&a Fra.tfyi.iM^iy, NwVoftk, and by NOSMO KING, Raieigh, N. C.Probtm Ed.itot14 Note.Readers interested in si'ew quadrilaterals will find solutions ofthe introductory theorem on pages 1026 and 1027 of the November 1973issue of the American Mathematical Monthly~solutions to Question B-5of the William Lowell Futnam Iiathema!ical Compet'tion held on December -,1972).301. [Spring 19731 Proposed by Neat Jacob&, &ionx, N. V.(Lorre=ted). One-fifteent!. can be expressed in "decimals" ir. manvways, for example, as .0421 in base ¥;ight or as ,013 in base five.*hat 'n any base n, the "decima-" for one-fifteenth will have n:, morethan feur recurring digits.Sotu,Uon by Vonn&Uy J. John&on,Fotce. Bae., Oluo...hrwhlajot, USAF, W~Lgd-t-PaAtmc~ A^':The representat is-.n '-'f 1 "15 as a "decimaJ " f;xpressi',n to base ). ,(n = 2, 3, 4, '' ,) either terminates or has a repeating block f :enz-.1, 2, or 4. T pr:ive th:s, 't is suffi~iento show that there i s r -peating block ¥. length 4. Also it is sufficient to show tha^ 1- :2, 3, 4;-', then 15 divide- n(n4 - 1) since ". n(n4 - l)¥'lr = : .-K = f a./nz with 0 2 a . n, then- - 2is seen to represent a number in "decimal" base ti wi" . :length 4 digits; andTcomplete the proof that 15 divid--p = 3 and for p = 5, p divide? e'"ther i ?f~cientto show forp = 3, 5 thatifa- -. .is a multiple of p: thu~., l4 - 1 = : ' -215.A hhoiue.d by RICHARD A. GIBES, - -' -FIGURE 6 FIGURE 7ado; BOB PRIELIPP, The. Uiztvml A -A&? .hot\}~~i noted the. ~tt-L'spti~."& . .

Pto6uhofi P/u-eL+p c^ited the Teachv~'~ !,lanuat 6o/i Excursions Int,-llathemat'vos by Beck, W.~ichefi and Cfiowe, pp, 271 -272. Gibbh no-tid tliegenvLoJL fiuuJU 6ound .in 8. 1.1. Stwatt, l.laanite.a.11 1962, p. 234.302. [Spring 19731 Pfiopmed by Vauid L. SU.vvman, We-it Loi Angde-i,CaH.6ofii'u.a and U.6te.d E.Ne.wnan, l.lu Aiplia VeUa Ffiatvt~wLy, New Yufik.A ta~~-~try i? hung on a wall so that its upper edge is a un.l*.. an?its lower edge b unit;, above the observer's eye-level.Show tha: inorder ti i.btain the mcct fav~rable view the observer should stand at adistance /ab from the waL:.I. Sotu^ion by R. Robire~on Rowe., Sac~.ame.nto, Catif,otiua.In Fig. 8, AB is the tapestry on wall ABC and tht? 2b:-erver'r i-y'-' :-at I a t a distance of IC = d from the wall.angle AIB = 9 should be a maximum.Since 1.'rt= AIC - BI,* = tanla/d - tanlb'd= tan 1 d(a + b)d"- abTor the most favorable view,'s less than go0, it :an be maximized by maximizing itstangent, and hence by maximizing its tangent divided bv (a - b).ting F(d) = d/(d2 + ab), we obtain Ft(d) = (ab d- )'(d- + abl2 = ,whence ab - d2 = ,, and d = ¥/ISComment by Pfiobiem EditofiLet-While this solution is correct in theory, ther.' are practical lim-itations. As C approaches B, b and d approach zero and 1 approaches 90'Thus, if a = 4 and b = .01, we find that d = .2 and even a myopic obser-ver would get a jaundiced view.:onstruct the cir-le thrwgh the top and the bc-tt~m pf the tapestrydnd tangent t. the eye-level line of the ..bserver (Fig. 9). The pointof crntact .f the circle wit:, the line is the optimum pcqition.For allpoints on the line :'ut~ide the circle, the tapestry subtends a smallerangle than at 'he rointis /ab.f ¥'-angency Hence the di-tance from the wallE. ANDERSON, Montana State U~uucuitq; ROBERT C . GEBHARDT, Hopatcong ,N. J.; JOHN M. HOWELL, Lct^fyiock, CaU&ofiiua; DONNELLY JOHNSON, W/u.gld-Patt&uon A^A Fome Bale, 0l~t.o; G. MAVRIGIAM, Youngh.town State U~uu~~lLty;BOB PRIELIPP, The. U~~iumLty 06 W-iiicoiu-in, O~Itko^h; DAN SCHOLTEN, WuteyanUihmLty. AAio hotved by the Pfiopoh&u UA-CII~ &if\vthe,tic ge.ome^iy.Pfiobiem EcLUtofi'ii Cotnmc~vtIt is hard to find an elementary calculus text that has not used;om

INDEX FOR VOLUME 5AVAILABLE ON REQUEST ONLYThe index to Volume 5 of the J o dwill be prepared for mailingwith the Fall 1974 issue, but only institutional subscribers (such aslibraries) will receive the index automatically.If you anticipate theneed for the index or would like to have one please let us know.WriteEditor, Pi I'iu Epsilon601 Elm Avenue, Room 423University of OklahomaNorman, Oklahoma 73069FRATERNITY KEY-PINS AVAILABLE?old key-pins are available at the National .,ffire at thespecial price of $5.0;:n::edStates.each, post paid to anywhere !n theBe. 4u~e to indicate. the. chaptvi into wfuch con wvie. hu.t-La-te.dand the. appfiohate. date. of, tile. JiH^-fct.cuLLon.Orders should be sent to:Ti tlu Epsilon, Inc.601 Elm Avenue, Room 423University of OklahomaNorman, Oklahoma 73069

YOUR BADGE - a triumph of sk ,:ed a"! - r "a -& Sa '.dcraftsmen isa steadfast and dynam c sti-rhi.' a -2-.i-3 -23Official Badge 1Official one piece keyOfficial one piece key-pinfOfficial three-piece keyOfficial three-piece key-pinWRITE FOR INSIGNIA PRICE LIST.OFFICIAL JEWELER TO PI MU EPSILONÂ¥