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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