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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

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