Luc TARTAR to John MACKEY, November 6, 2009
Your Putnam set 11.
Your problem 1: 3n = 6 (mod 10) means n = 2 (mod 10), so that 3n = 3(10m + 2) = 30m + 6, and it is a
five digit if and only if 10 000 ≤ 30m + 6 < 99 999, so that 10 000 ≤ 30m < 100 000, or 1 000 ≤ 3m < 10 000,
which is equivalent to 1 002 ≤ 3m ≤ 9 999, or 334 ≤ m ≤ 3 333. There are 3 000 such numbers.
Your problem 2: The equation of the tangent at the curve y = x3 at the point (a, a3 ) is y − a3 = 3a2 (x − a),
i.e. 2a3 − 3a2 x + y = 0. A cubic equation which has only two solutions is such that one solution satisfies the
derived equation, 6a2 − 6a x = 0, so that either a = 0 or a = x. The case a = 0 gives y = 0, i.e. the x axis,
but one must exclude x = 0, because a = 0 is then a triple root, and there is only one solution. The case
a = x gives y = x3 , i.e. the given curve, but one must also exclude x = 0.
Your problem 3 (Putnam 1968-A1): Prove
1
x4 (1 − x)4
Z
22
−π = dx.
7 0 1 + x2
Hint: One applies the usual method for integrating rational fractions. x4 (1−x)4 = x8 −4x7 +6x6 −4x5 +x4 =
R1 R1
(x2 + 1) (x6 − 4x5 + 5x4 − 4x2 + 4) − 4, so that the integral is 0 (x6 − 4x5 + 5x4 − 4x2 + 4) dx − 4 0 x2dx
+1 =
1 = 22 − π.
1 4 5 4
7 − 6 + 5 − 3 + 4 − 4 arctan x 0 7
√
Your problem 4 (Putnam 2001-B3): For any positive integer n, let hni denote the closest integer to n.
Evaluate
∞
X 2hni + 2−hni
.
n=1
2n
√
I had not written a solution before: hni = m means m − 12 < n < m + 21 , or after taking the squares
m2 − m + 14 < n < m2 + m + 41 , and since n is a integer, it is equivalent to m2 − m + 1 ≤ n ≤ m2 + m. For
Pn=m2 +m Pn=b
m ≥ 0, one then has a group of terms (2m + 2−m ) n=m2 −m+1 21n , and since for a < b the sum n=a 21n is
1 1
is (2m + 2−m ) 2m21−m − 2m21+m = 2m21−2m − 2m21+2m . One then has a telescopic
2a−1 − 2b , the group of terms
1 1 1 1 1 1 1
sum: (1 − 1) + 2−1·1 − 21·3 + 1 − 22·4 + 21·3 − 23·5 + 22·4 − 24·6 + . . . so that the sum for m even is 1
1
and the sum for m odd is 2−1·1 = 2, so that the total sum is 3.
Your problem 5 (Putnam 1980-B4): Let A1 , A2 , . . . , A1066 be subsets of a finite set X such that |Ai | > 12 |X|
for 1 ≤ i ≤ 1066. Prove there exist ten elements x1 , . . . , x10 of X such that every Ai contains one of
x1 , . . . , x10 .
[Here |S| means the number of elements in the set S.] P
Hint: Suppose |X| = n, and one considers 2m + 2 set Aj , then because j |Aj | > (m + 1)n one deduces by
the pigeon hole principle that there exists z belonging to m + 2 of the Aj ,P which one puts aside and one is
left with m sets Aj ; similarly, if one considers 2m + 1 set Aj , then because j |Aj | > m + 12 n one deduces
that there exists z belonging to m + 1 of the Aj , which one puts aside and one is left with m sets Aj .
One starts with 1066 sets and one finds x1 belonging to 534 of them, and one is left with 532 sets; there
exists x2 belonging to 267 of them (and it does not matter if x2 coincides with x1 ) and one is left with 265
sets; there exists x3 belonging to 133 of them and one is left with 132 sets; there exists x4 belonging to 67
of them and one is left with 65 sets; there exists x5 belonging to 33 of them and one is left with 32 sets;
there exists x6 belonging to 17 of them and one is left with 15 sets; there exists x7 belonging to 8 of them
and one is left with 7 sets; there exists x8 belonging to 4 of them and one is left with 3 sets; there exists x9
belonging to 2 of them and one is left with 1 set, in which one picks x10 .
Actually, 1066 could be replaced by N with 766 < N ≤ 1534, and more precisely, if 3.2k−2 − 2 < N ≤
3.2k−1 − 2, and one has N sets with |Ai | > 12 |X|, then one can finds x1 , . . . , xk of X such that every Ai
contains one of x1 , . . . , xk ; this comes from the fact that if 2m + 2 = 3.2k−1 − 2, then m = 3.2k−2 − 2, and
that k = 1 corresponds to N = 1.
