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STOP Early Journal Content on JSTOR, Free to Anyone in the World This article is one of nearly 500,000 scholarly works digitized and made freely available to everyone in the world by JSTOR. Known as the Early Journal Content, this set of works include research articles, news, letters, and other writings published in more than 200 of the oldest leading academic journals. The works date from the mid-seventeenth to the early twentieth centuries. We encourage people to read and share the Early Journal Content openly and to tell others that this resource exists. People may post this content online or redistribute in any way for non-commercial purposes. Read more about Early Journal Content at http://about.jstor.org/participate-jstor/individuals/early- journal-content . JSTOR is a digital library of academic journals, books, and primary source objects. JSTOR helps people discover, use, and build upon a wide range of content through a powerful research and teaching platform, and preserves this content for future generations. JSTOR is part of ITHAKA, a not-for-profit organization that also includes Ithaka S+R and Portico. For more information about JSTOR, please contact support@jstor.org. 130 PROBLEMS AND SOLUTIONS. from A and is the angle which the string makes with the horizontal at the given point. It can be easily shown that tan d = sinh x/c, so that sin 6 = tanh x/c,. and hence T = ws coth x/c. At the point B we have s = c sinh d/2c and T — P, say. Since the hook is smooth P = W and we have wl = wc sinh d/2c coth d/2c = wc cosh d/2c. Hence d_ I = c cosh d/2c and \L = Z + s = c(cosh d/2c + sinh d/2c) = ceK We must now find the value of c which makes fi a minimum. From the equation D C (^L) = we obtain jL d i. d 2c (-')-* d or Hence \L — d/2 • e = de/2 or L = de. Also solved by J. B. Smith. NUMBER THEORY. 187. Proposed by E. T. bell, New York, N. Y. If m is any integer, P the product of all the distinct prime factors of m and X their number,, and if N{x) denote the number of divisors of x, then &2N(d)N(m/d) = N(m)N{Pm)N(Phn), where the sign of summation extends over all the divisors d of m. Solution by Thomas E. Mason, Indiana University. Since m has X distinct prime factors it can be written m where the p's are distinct primes. By a well-known formula we have N(m) = ( ai + 1)(« 2 + 1) • • • K + 1), N(Pm) = {a x + 2) • • • (a K + 2), iV(P 2 m) = («! + 3) • • • (a K + 3). Any divisor d of m can be written in the form d = ?r W 3 • • • P^- N(d)= (n+lXrs+l)... (r x +l), N(m/d) = (ax - n + l)(a 2 - r 2 + 1) • • • (a K - r k + 1). pfWpf • • • p A a \ (1) Then and PROBLEMS AND SOLUTIONS. 131 Now we have WN(d)N(m/d) = Q" £ ( ai _ fl +l)(a 2 -r 2 +l) • • • (a A -r A +l)(n+l)(r 2 +l) fj=0 ••■(r*+l) = 6 A J (ax - n + l)(n + 1) • • • (a A - r A + l)( fA + 1) (2) = 6 D [ ai (n + 1) - (r x 2 - 1)] ■ 6 £ [«2(r 2 + 1) - (r S 2 - 1)] ri=0 r2=0 ■8Ek(r»+l)-W-l)]. E The indicated summations can be readily performed. We have 2^ 0; + 1) = s h «» + 1 = o and f. ( , u ai(a i +l)(2a i +l) fa + l)(g f + 2)(2a< - 3) £0^ ; ~ 6 «» 1 - 6 Hence • £6*fr<+ 1) - W- Dl = 6[ ( ,, ( <*+ 1 f + 2 > _ ( a .+ l)(c+2)(2«,-3) j = (a«+l)(a,-+2)(a«+3). Substituting for these sums in the last member of (2) and making use of (1), we have V2N(d)N(m/d) = (at + l)( ai + 2){a x + 3)(« 2 + 1)(« 2 + 2)(« 2 + 3) ••• K+l)(« A +2)(« A +3) = N(m)N(Pm)N(P ,i m). 197. Proposed by E. T. BELL, Seattle, Washington. Show that in the expansion of 1 + z + z 2 + • • • + z*- 1 , (1 - z)"- 1 where p is a prime, the coefficients of the various powers of z are divisible by p. [Eisenstein, Crelle, t. 27, p. 282.] Solution by B. F. Yanney, University of Wooster. The given expression is evidently equal to (1 — s p )/(l — z) p — 1, which may be put in the form (1 — z p )(l — z)~ p — 1. Expanding the second factor of the first term and noticing that, since p is prime, the coefficient of each term