Luc TARTAR to John MACKEY, September 29, 2009

For your problem set 6 of today, September 29, I had already written solutions in the preceding years.
Your problem 1 (Putnam 1969-B1): Let n be a positive integer such that n + 1 is divisible by 24. Prove that
the sum of all divisors of n is divisible by 24.
Hint: If d1 is a divisor of n, then n = d1 d2 and d2 is also a divisor of n and if one shows that d1 + d2 is
automatically a multiple of 24, and that n is not a square (so that d1 is different from d2 ), then the sum
of all divisors of n will be a multiple of 24. As n = 23 (mod 24), one has n = 2 (mod 3), so that n is
not a square (since squares are 0 or 1 modulo 3). Either d1 = 1 (mod 3), so that d2 = 2 (mod 3), or
d1 = 2 (mod 3), so that d2 = 1 (mod 3): in both cases d1 + d2 = 0 (mod 3). Similarly n = 7 (mod 8),
so that d1 must be congruent to 1, 3, 5 or 7 (mod 8), and in order to have d1 d2 = 7 (mod 8), d2 must be
congruent to respectively 7, 5, 3, or 1 (mod 8), and in each of these cases one has d1 + d2 = 0 (mod 8).
Then, d1 + d2 is a multiple of 3 and a multiple of 8, so that it is a multiple of 24.
Your problem 2 (Putnam 1985-A1): Determine, with proof, the number of ordered triples (A1 , A2 , A3 ) of
sets which have the property that
(i) A1 ∪ A2 ∪ A3 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},
and
(ii) A1 ∩ A2 ∩ A3 = ∅,
where ∅ denotes the empty set. Express the answer in the form 2a 3b 5c 7d , where a, b, c and d are nonnegative
integers.
Hint: Each of the ten integers i ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} should be put in at least one of the three sets
A1 , A2 , A3 , but not in the three sets, and there are 6 possibilities for that (out of 23 possibilities, reject that
where i is put in none or in the three sets). The desired number is then 610 = 210 310 .
Your problem 3 (Putnam 1970-A4): Given a sequence xn , n = 1, 2, · · · such that

lim xn − xn−2 = 0.
n→∞

Prove that
xn − xn−1
lim = 0.
n→∞ n
Hint: The hypothesis is made up of two unrelated hypothesis, namely limn→∞ x2n − x2n−2 = 0 concerning
even indices, and limn→∞ x2n+1 − x2n−1 = 0 concerning odd indices, but the conclusion involves successive
terms. The reason is that one must show that if a sequence yn is such that limn→∞ yn − yn−1 = 0, then one
has limn→∞ ynn = 0; then, choosing yn = x2n shows that limn→∞ xn2n = 0, and choosing yn = x2n+1 shows
that limn→∞ x2n+1
n = 0, and a consequence is that limn→∞ xnn = limn→∞ xn−1 n = 0.
For ε > 0, one has |yn+1 − yn | ≤ ε for n ≥ n(ε), so that |yn(ε)+p − yn(ε) | ≤ p ε; from |yn | ≤ |yn(ε) | + ε n
|y |+ε n
for n ≥ n(ε), one deduces that lim supn→∞ |ynn | ≤ limn→∞ n(ε)n = ε, and then letting ε tend to 0 shows
that lim supn→∞ n = 0, and since n ≥ 0, one deduces that limn→∞ |ynn | = 0.
|yn | |yn |

Your problem 4 (Putnam 1970-B6): A quadrilateral which can be inscribed in a circle is said to be inscribable
or cyclic. A quadrilateral which can be circumscribed to √ a circle is said to be circumscribable. Show that if
a circumscribable quadrilateral of sides a, b, c, d has area a b c d, then it is also inscribable.
Hint: Let α be the angle whose sides are a, b and let β be the angle whose sides are c, d; if one shows that
α + β = π, then it proves that the quadrilateral is inscribable.
One has a2 + b2 − 2a b cos α = c2 + d2 − 2c d cos β, as it is the square of the diagonal opposite to the
angles α or β. Since the two tangents to the circle from any point outside it have the same length, one
deduces that any circumscribable quadrilateral satisfies a + c = b + d; one then has (a − b)2 = (c − d)2 , hence
a b (1 − cos α) = c d (1 − cos β). √
The area of the quadrilateral is a b sin
2
α
+ c d sin
2
β
, which is assumed to be a b c d, so that taking the square
gives 4a b c d = a2 b2 sin2 α + c2 d2 sin2 β + 2a b c d sin α sin β, but since a2 b2 sin2 α = a b (1 + cos α) c d (1 − cos β)
and c2 d2 sin2 β = a b (1−cos α) c d (1+cos β), one obtains 4 = 2−2 cos α cos β+2 sin α sin β = 2−2 cos(α+β),
i.e. α + β = π.

1
Your problem 5 (Putnam 1971-B2): Let F (x) be a real valued function defined for all real x except for x = 0
and x = 1 and satisfying the functional equation F (x) + F {(x − 1)/x} = 1 + x. Find all functions F (x)
satisfying these conditions.
−1 −1
Hint: Let ϕ(x) = x−1 x−1



x . Then ϕ ◦ ϕ(x) = ϕ x
= x−1 , and ϕ ◦ ϕ ◦ ϕ(x) = ϕ x−1 = x. Then, from
F (x)+F ϕ(x) = 1+x, one deduces that F ϕ(x) +F ϕ◦ϕ(x) = 1+ϕ(x) = 2x−1




x and F ϕ◦ϕ(x) +F (x) =
x−2 2x−1 x−2 x3 −x2 −1
1 + ϕ ◦ ϕ(x) = x−1 , so that 2F (x) = 1 + x − x + x−1 , i.e. F (x) = 2x(x−1) .