Math 17 Fall 2015
Serin Hong
PUTNAM 2016 SOLUTIONS
Problem A1. Find the smallest positive integer j such that for every polynomial p(x) with integer coefficients
and for every integer k, the integer
p(j) (k) =
dj
p(x)|x=k
dxj
(the j-th derivative of p(x) at k) is divisible by 2016.
Solution:
We claim that j = 8 is the smallest such integer.
Write P (x) = a0 + a1 x + a2 x2 + · · · + ad xd . Then
P (j) (k) =
d
X
i(i − 1)(i − 2) · · · (i − j + 1)ai k j−i .
i=j
It is easy to observe that the product of 8 consecutive inters is divisible by 25 , 32 and 7, and therefore by
2016 = 25 · 32 · 7. Hence if j ≥ 8, each term in the above sum is divisible by 2016, implying that p(j) (k) is
divisible by 2016.
For j < 8, take p(x) = xj . Then p(j) (k) = j! is not divisible by 2016.
Problem A2. Given a positive integer n, let M (n) be the largest integer m such that
m
m−1
>
.
n−1
n
Evaluate
lim
n→∞
M (n)
.
n
Solution:
The given inequality is equivalent to mn > (m−n+1)(m−n), or m2 −(3n−1)m+n2 −n < 0. Considering
this as a quadratic inequality in m, we obtain
(3n − 1) −
√
√
5n2 − 2n + 1
(3n − 1) + 5n2 − 2n + 1
<m<
2
2
which yields
M (n) = b
(3n − 1) +
√
5n2 − 2n + 1
c
2
Math 17 Fall 2015
Serin Hong
where b·c is the Guass floor function. Then we have
(3n − 1) +
√
√
(3n − 1) + 5n2 − 2n + 1
5n2 − 2n + 1
− 1 ≤ M (n) ≤
.
2
2
Dividing this inequality by n and taking the limit yields
√
M (n)
3+ 5
lim
=
.
n→∞
n
2
Problem A3. Suppose that f is a function from R to R such that
f (x) + f
1
1−
x
= arctan x
for all real x 6= 0. (As usual, y = arctan x means −π/2 < y < π/2 and tan y = x.) Find
1
Z
f (x)dx.
0
Solution :
The given functional equation is
f (x) + f
Substituiting 1 −
1
x
1−
= arctan x
for x 6= 0.
1
1
and
to (1) respectively yield
x
1−x
1
1
1
f 1−
+f
= arctan 1 −
x
1−x
x
1
1
+ f (x) = arctan
f
1−x
1−x
(1)
for x 6= 0, 1,
(2)
for x 6= 0, 1.
(3)
Then (1) - (2) + (3) yields
2f (x) = arctan x + arctan
1
1−x
1
− arctan 1 −
x
for x 6= 0, 1.
and hence
Z
2
1
Z
f (x)dx =
0
1
Z
arctan x dx +
0
1
arctan
0
1
1−x
Z
dx −
0
1
1
arctan 1 −
x
dx.
(4)
Math 17 Fall 2015
Serin Hong
To calculate the first two integrals in (4), we simply note that for x ∈ (0, 1) we have arctan
π
− arctan(1 − x), which yields
2
Z 1
Z 1
Z 1
π
π
1
arctan x dx.
dx =
− arctan(1 − x) dx = −
arctan
1−x
2
2
0
0
0
For the third integral in (4), we set y = arctan
1
1−x
=
1
1
. Then we have x =
, so using the inverse
1−x
1 − tan y
function trick we obtain
1
Z
arctan
0
1
1−x
Z
0
=−
−π/2
dy
.
1 − tan y
On the other hand, we have
π
1 + tan y
2
=
+ 1 = tan
+ y + 1.
1 − tan y
1 − tan y
4
Hence we calculate the integral by
Z
0
−π/2
dy
1
=
1 − tan y
2
Z
0
tan
Z
1 π/4
π
π
+ y + 1 dy =
tan y dy + = .
4
2 −π/4
4
4
π
−π/2
By (4) we get
Z
1
f (x)dx =
0
3π
.
8
Problem A 4. Consider a (2m − 1) × (2n − 1) rectangular region, where m and n are integers such that
m, n > 4. This region is to be tiled using tiles of the two types shown:
(The dotted lines divide the tiles into 1 × 1 squares.) The tiles may be rotated and reflected, as long as
their sides are parallel to the sides of the rectangular region. They must all fit within the region, and they
must cover it without overlapping.
What is the minimum number of tiles required to tile the region?
Solution:
We claim that the minimum number of tiles required is mn.
Consider the following coloring of the region:
Math 17 Fall 2015
Serin Hong
···
···
···
..
.
..
.
..
.
..
.
..
.
..
.
Note that each tile can cover at most one black square. There are mn black squares, so we need at least mn
tiles to cover the region.
Now it remains to prove that we can cover the region with mn tiles. We prove this by induction on m
and n. When m = n = 4, we have the following covering:
The inductive step follows from the following covering of 2 × m (or 2 × n) rectangle:
···
···
Problem A 5. Suppose that G is a finite group generated by the two elements g and h, where the order of
g is odd. Show that every element of G can be written in the form
g m1 hn1 g m2 hn2 · · · g mr hmr
with 1 ≤ r ≤ |G| and m1 , n1 , m2 , n2 , · · · , mr , nr ∈ {−1, 1}. (Here |G| is the number of elements of G.)
Math 17 Fall 2015
Serin Hong
Solution:
We call an expression of an element of G as stated in the problem a long form, and define r to be its
length. For example, ghg −1 h−1 is a long form with length 2.
Let H be the cyclic subgroup of G generated by ghg −1 h−1 . Let s be the smallest integer such that g s ∈ H,
which must be odd as it divides the order of G. Note that we have (gh)H = (hg)H, and therefore
(xy)H = (yx)H
for any x, y ∈ G.
(5)
In particular, we can choose the coset representatives to be {g i hj : 0 ≤ i ≤ s − 1, 0 ≤ j ≤ t − 1} where t is
the smallest integer such that ht ∈ g u (H) for some u. It suffices to prove that every element in (g i hj )H can
be expressed in a long form.
Note that if s = t = 1, we have G = H. Since every element in H are clearly in a long form, the assertion
is true. We will now assume that st ≥ 2.
Using (5) we find
(g i )H =



(g s−i )H = (g −1 hg −1 h−1 )(s−i)/2 H
if i is odd,


(ghgh−1 )i/2 H
if i is even.
j
(hj )H = (ghgh−1 )(s−1)/2 gh H.
Hence we have
(g i hj )H =

j


(g −1 hg −1 h−1 )(s−i)/2 (ghgh−1 )(s−1)/2 gh H


(ghgh−1 )i/2
j
(ghgh−1 )(s−1)/2 gh H
if i is odd,
.
if i is even.
In other words, every element x in (g i hj )H can be written in the form
x=

j


(g −1 hg −1 h−1 )(s−i)/2 (ghgh−1 )(s−1)/2 gh (ghg −1 h−1 )k


(ghgh−1 )i/2
j
(ghgh−1 )(s−1)/2 gh (ghg −1 h−1 )k
if i is odd,
if i is even.
for some integer k ≤ |H| − 1. Now it remains to check that these expressions have length at most |G|. If i
is odd, the length is (s − i) + (s − 1)j + 2k ≤ s + (s − 1)(t − 1) + 2k ≤ st + 2(|H| − 1) ≤ st|H| = |G|, where
the last inequality follows from st ≥ 2. Similarly, we can show that the length is at most |G| when i is even.
Math 17 Fall 2015
Serin Hong
Problem A 6. Find the smallest constant C such that for every real polynomial P (x) of degree 3 that has
a root in the interval [0, 1],
Z
1
|P (x)|dx ≤ C max |P (x)|.
0
Solution:
No idea.
x∈[0,1]
Math 17 Fall 2015
Serin Hong
Problem B1. Let x0 , x1 , x2 , · · · be the sequence such that x0 = 1 and for n ≥ 0,
xn+1 = ln(exn − xn )
(as usual, the function ln is the natural logarithm). Show that the infinite series
x0 + x1 + x2 + · · ·
converges and find its sum.
Solution:
Since ex − x > 1 for all x > 0, an easy induction shows that xn > 0 for all n ≥ 0. On the other hand, the
given recursive equation can be written as
xn = exn+1 − exn
which yields
x0 + x1 + · · · + xn = ex0 − exn+1 = e − exn+1
for all n ≥ 0.
(6)
In particular, the partial sum x0 + x1 + · · · + xn is strictly increasing and bounded above, so the infinite
series x0 + x1 + · · · converges. This implies that xn → 0 as n → ∞, and therefore (6) yields
x0 + x1 + · · · = e − 1.
Problem B2. Define a positive integer n to be squarish if either n is itself a perfect square or the distance
from n to the nearest square is a perfect square. For example, 2016 is squarish, because the nearest perfect
square to 2016 is 452 = 2025 and 2025 − 2016 = 9 is a perfect square. (Of the positive integers between 1
and 10, only 6 and 7 are not squarish.)
For a positive integer N , let S(N ) be the number of squarish integers between 1 and N , inclusive. Find
positive constants α and β such that
S(N )
=β
N →∞ N α
lim
or show that no such constants exist.
Solution:
We claim that α = 3/4 and β = 4/3.
We use the little-o notation. For example, we say that a given function h(n) is o(nc ) if lim
n→∞
h(n)
= 0.
nc
Math 17 Fall 2015
Serin Hong
Let k 2 be the nearest perfect square to N , i.e., (k − 1/2)2 ≤ N ≤ (k + 1/2)2 . For each positive integer j,
√
√
the number of squarish integers in the interval [(j − 1/2)2 , (j + 1/2)2 ] is b j − 1c + b jc + 1. Hence
k−1
X
S(N ) =
(b
p
j − 1c + b
p
jc + 1) + j=0
where is the number of squarish integers in the interval [(k − 1/2)2 , N ]. Note that
√
√
√
0 ≤ ≤ b k − 1c + b kc + 1 ≤ 2 k + 1.
Hence we may write
S(N ) =
k−1
X
p
p
√
√
(b j − 1c + b jc + 1) + b k − 1c + ( − b k − 1c)
j=0
=2
k−1
X
b
p
√
jc + (k − 1) + ( − b k − 1c)
j=1
=2
k−1
X
b
p
jc + o(N 4/3 ).
(7)
j=1
Now let l be the positive integer such that l2 ≤ k < (l + 1)2 . Then
k−1
X
b
p
jc = 1 · (22 − 12 ) + 2 · (32 − 22 ) + · + (l − 1)(l2 − (l − 1)2 ) + l(k − l2 )
j=1
=
k−1
X
j((j + 1)2 − j 2 ) + l(k − l2 )
j=1
=
k−1
X
j(2j + 1) + l(k − l2 )
j=1
2(l − 1)l(2l − 1) (l − 1)l
+
+ l(k − l2 ).
3
2
q√
q√
√
√
Note that l ≤ k ≤
N + 1/2 and l > k − 1 ≥
N − 1/2 − 1, so l = N 1/4 + o(N 1/4 ). On the other
=
hand, we have 0 ≤ k − l2 < (l + 1)2 − l2 = 2l + 1. Hence we can write
k−1
X
b
j=1
p
2
jc = N 3/4 + o(N 3/4 ).
3
Thus we deduce from (7) that
S(N ) =
which yields α = 3/4 and β = 4/3 as desired.
4 3/4
N
+ o(N 3/4 ),
3
Math 17 Fall 2015
Serin Hong
Problem B 3. Suppose that S is a finite set of points in the plane such that the area of triangle ∆ABC
is at most 1 whenever A, B, C are in S. Show that there exists a triangle of area 4 that (together with its
interior) covers the set S.
Solution:
Choose P1 , P2 , P3 ∈ S with the maximum area of ∆P1 P2 P3 . Let l1 be the line passing through P1 which
is parallel to P2 P3 , and m1 be the reflection of l1 to P2 P3 . We similarly define the lines l2 , l3 , m2 , m3 .
m3
m2
l3
P1
l1
Q3
Q2
P2
P3
m1
Q1
l2
Let Q1 be the intersection of l2 and l3 , and similarly define Q2 and Q3 . Then one easily sees that mi passes
through Qi for i = 1, 2, 3 as in the figure. The three points P1 , P2 , P3 are the midpoints of the sides of the
triangle ∆Q1 Q2 Q3 , so we have
Area(∆Q1 Q2 Q3 ) = 4Area(∆P1 P2 P3 ).
The area of the triangle ∆Q1 Q2 Q3 is at most 4 as the area of the triangle ∆P1 P2 P3 is at most 1.
Let P be a point in S. Since Area(∆P P2 P3 ) ≥ Area(∆P1 P2 P3 ), P must lie between l1 and m1 . Similarly,
P must lie between l2 and m2 , and also between l3 and m3 . Hence P must lie in the triangle ∆Q1 Q2 Q3 .
This implies that the triangle the set S is covered by the triangle ∆Q1 Q2 Q3 , whose area is at most 4 as seen
above.
Problem B 4. Let A be a 2n × 2n matrix, with entries chosen independently at random. Every entry is
chosen to be either 0 or 1, each with probability 1/2. Find the expected value of det(A − At ) (as a function
of n), where At is the transpose of A.
Math 17 Fall 2015
Serin Hong
Solution:
Let B = A − At . Let ai,j (resp. bi,j ) denote the (i, j)-entry of A (resp. B). We know that B is
anti-symmetric, i.e., bi,j = ai,j − aj,i for 1 ≤ i, j ≤ 2n. In particular,
bi,i = 0
for i = 1, 2, · · · , 2n,
and if i 6= j, the probability distribution of bi,j is given as follows:
bi,j =




0




1






−1
with probability 1/2
(8)
with probability 1/4
with probability 1/4.
Furthermore, two random variables bi,j and bi0 ,j 0 are independent unless i = j 0 and j = i0 .
The expected value of det(B) is given by
!
X
E(det(B)) = E
sgn(σ)b1,σ(1) b2,σ(2) · · · b2n,σ(2n)
σ∈S2n
=
X
sgn(σ)E b1,σ(1) b2,σ(2) · · · b2n,σ(2n) .
(9)
σ∈S2n
where S2n is the group of permutations of {1, 2, · · · , 2n}.
We claim that E b1,σ(1) b2,σ(2) · · · b2n,σ(2n) = 0 unless σ is a product of n transpositions, i.e. every element
in {1, 2, · · · , 2n} has order 2 under the permutation σ. Suppose that there exists an element i ∈ {1, 2, · · · , 2n}
whose order under σ is not equal to 2. If i is a fixed point (i.e., the order is 1) then bi,σ(i) = 0 and therefore
E b1,σ(1) b2,σ(2) · · · b2n,σ(2n) = 0. Now suppose that the order of i under σ is greater than 2. Then σ 2 (i) 6= i,
so bi,σ(i) is independent of bj,σ(j) for j 6= i. This yields
E b1,σ(1) b2,σ(2) · · · b2n,σ(2n) = E b1,σ(1) b2,σ(2) · · · bi−1,σ(i−1) bi+1,σ(i+1) · · · b2n,σ(2n) E bi,σ(i) = 0
where the second equality follows from the fact that E bi,σ(i) = 0. Hence we establish the claim.
Now we assume that σ is a product of n transpositions. Then we may write
σ = (i1 j1 )(i2 j2 ) · · · (in jn ).
Observe that bi1 ,j1 , bi2 ,j2 , · · · , bin ,jn are independent variables. We also have E(−b2ik ,jk ) = −
1
by the prob2
ability distribution given in (8) for k = 1, 2, · · · , n. Hence we calculate
n
n
n
Y
Y
Y
1 n
E b1,σ(1) b2,σ(2) · · · b2n,σ(2n) = E
bik ,jk bjk ,ik = E
−b2ik ,jk =
E − b2ik ,jk = −
2
k=1
k=1
k=1
Math 17 Fall 2015
Serin Hong
where the second equality follows from bjk ,ik = −bik ,jk . On the other hand, sgn(σ) = (−1)n as σ is a product
of n transpositions, so we find
n
1
1
sgn(σ)E b1,σ(1) b2,σ(2) · · · b2n,σ(2n) = (−1)n · −
= n.
2
2
N
where N is the number of permutations in S2n which are a
2n
product of n transpositions. One easily finds that
We deduce from (9) that E(det(B)) =
N = (2n − 1) · (2n − 3) · · · 3 · 1 =
(2n)!
(2n)!
= n
,
(2n)(2n − 2) · · · 4 · 2
2 · n!
so we conclude that
E(det(B)) =
(2n)!
.
22n · n!
Problem B5. Find all functions f from (1, ∞) to (1, ∞) with the following property:
if x, y ∈ (1, ∞) and x2 ≤ y ≤ x3 , then f (x)2 ≤ f (y) ≤ f (x)3 .
Solution:
No idea.
Problem B6. Evaluate
∞
∞
X
(−1)k−1 X
k=1
k
n=0
1
.
+1
k2n
Solution:
We rewrite the given sum as
∞
∞
X
(−1)k−1 X
k=1
k
∞
∞
X X (−1)k−1
1
1
=
· n
.
n
k2
+
1
k
k2
+1
n=0
n=0
(10)
k=1
Note that every integer m > 1 can be uniquely written in the form m = r · 2s + 1 where r, s are nonnegative
integers with r odd. We consider the terms in (10) such that k2n + 1 = m. Such terms appear when
k = r · 2s−n and n = 0, 1, · · · , s. For such terms, we have (−1)k = 1 if and only if n = s. In other words,




−
1
1
·
1
(−1)k−1
r · 2s−n m
· n
=
k
k2 + 1 


 1· 1
r m
if 0 ≤ n ≤ s − 1
if n = s.
Math 17 Fall 2015
Serin Hong
Hence the sum of these terms is
s−1
X
1
1 1
1
1
·
·
−
=
r m n=0 r · 2s−n m
rm
1−
s−1
X
1
2s−n
n=0
!
=
1
1 1
1
1
1
= s
=
=
− .
rm 2s
r2 · m
(m − 1)m
m−1 m
Thus the sum in (10) can be computed by
X
m>1
1
1
−
m−1 m
= 1.