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PRIMES Math Problem Set: Solutions
Evan Chen
PRIMES 2021
Solution to General Math Problems
Problem G
A polynomial f (x) has complex coefficients. It turns out that f (x) · f ′(x) is a degree five polynomial whose x^5 , x^4 , x^1 , x^0 coefficients are respectively 3, 10, 25,
- Determine the polynomial f.
Solution
The answers are
f (x) = x^3 + 2x^2 + 3x + 4,
f (x) = x^3 + 2x^2 +
x +
f (x) = x^3 + 2x^2 +
x +
and the negations of these (for a total of six possible answers). Notice that we have
2 f (x)f ′(x) = 6x^5 + 20x^4 + Fx^3 + Fx^2 + 50x^1 + 24
where F represents coefficients that are not known. The left-hand side is the derivative of the polynomial f (x)^2 , so it follows (by integrating both sides) that
f (x)^2 = x^6 + 4x^5 + Fx^4 + Fx^3 + 25x^2 + 24x + F.
Apparently, f is a cubic polynomial. By replacing f with −f if necessary, we may as well assume f is monic. So we are seeking constants a, b, c such that
(x^3 + ax^2 + bx + c)^2 = x^6 + 4x^5 + Fx^4 + Fx^3 + 25x^2 + 12x + F.
We therefore get the system of equations
2 a = 4, 2 ac + b^2 = 25, 2 bc = 24.
Evidently a = 2, and now solving the resulting cubic equation gives (b, c) = (3, 4) as well
as the two solutions (b, c) =
− 3 ± √ 73 2 ,^
9 ± 3 √ 73 8
. This gives the solutions claimed.
Problem G
Scientists have found a vaccine that produces undesirable side effects with probabil- ity p. Initially, the number p is distributed uniformly across the interval [0, 0 .1]. To test the vaccine, the scientists test the vaccine on 148374 volunteers and find that no one experiences adverse side effects. Find the smallest real number λ such that the scientists can assert p < λ with probability at least 95%. Round your answer to four significant figures.
Solution
Let n = 148374 for brevity. By Bayes’ theorem, we require λ to satisfy
- 95 ≤ P(p < λ)
=
∫ (^) λ 0 (1^ −^ p)
n (^) dp ∫ (^0). 1 0 (1^ −^ p) n (^) dp
1 n+1 (1^ −^ (1^ −^ λ)
n+1) 1 n+1 (1^ −^ (1^ −^0 .1) n+1)
= 1 − (1 − λ)n+ 1 − 0. 9 n+^
Using exact methods will give λ = 2. 019 · 10 −^5 as the optimal choice. Actually, this can be approximated very closely by hand by simply commenting the denominator is very nearly 1 for large n; so essentially we want λ ≈ 1 − 0. 051 /(n+1)^ which gives the same approximation above.
Problem G
We roll a fair six-sided die and let s 1 be the result of the roll. Then, we roll s 1 fair six-sided dice and let s 2 be the sum of the rolls. Then, we roll s 2 fair six-sided dice and let s 3 be the sum of the rolls. The process continues to generate an infinite sequence (s 1 , s 2 ,... ).
(a) Find the probability that 3 appears in the sequence.
(b) Find the expected value of sn, for each integer n.
(c) We say the sequence grows exponentially if there exists a constant c > 1 such that sn > cn^ for all sufficiently large integers n. Does the sequence grow exponentially almost surely?
Solution
(a): The sequence is nondecreasing, so we can solve this by considering just a finite number of states. Let a be the probability of achieving 3 from sn if sn = 1, and let b be the probability of achieving 3 from sn = 2. Then it follows that
a =
· a +
· b +
b =
· b +
Solving gives b = 2/35 and a = 37/175, so the answer is 37/175. (b): sn = (7/2)n^ by induction. Indeed, sn+1 in general is equal to the sum of sn independent dice rolls, and each dice roll contributes 7/2. To make this rigorous we may write sn+1 =
k≥ 1
Xk · (^1) sn>k
where Xk is a dice roll, and (^1) sn>k is the relevant indicator variable. Then by taking linearity of expectation we get
E[sn+1] =
k≥ 1
· P(sn > k) =
k≥ 1
P(sn > k) =
E[sn]
proving the claim. (c): Yes (and the bounds are quite weak). Notice that sn+1 ≥ 32 sn holds as long as at least half the dice roll greater than 1, so it holds with probability at least 12 , say. Now by the central limit theorem, if we take c = 3
3 /2, then for sufficiently large N , the probability that sn < cn^ = (3/2)n/^3 decays exponentially in n. Taking a union bound across large enough N will give a total probability less than 1, as desired.
Problem G
For each positive integer n ≥ 4, find all positive real numbers a 1 , a 2 ,... , an such that a^2 i = 19ai+1 + 20ai+2 + 21ai+ holds for all i = 1,... , n with indices taken modulo n.
Solution
The answer is that all numbers must be equal to 60 (which works). The largest number is at most 60, since if M is maximal, then M 2 ≤ 60 M. The smallest number is at least 60, since if m is minimal, then m^2 ≥ 60 m. So all the numbers are equal to 60.
Solution to Advanced Math Problems
Problem M
For positive integers n, find a closed form for ∑
a+b+c+d=n a,b,c,d≥ 0
2 a+2b+3c+4d
in terms of n. Possible hint: use generating functions.
Solution
We use generating functions. Because (^1) −^12 X =
k 2 kXk, it follows the desired answer
is equivalent to the coefficient of xn^ in
F (X) =
1 − 2 X
1 − 4 X
1 − 8 X
1 − 16 X
We may express F (X) using partial fractions as
− 1 / 21 1 − 2 X
1 − 4 X
1 − 8 X
1 − 16 X
This gives the answer of
− 1 21 · 2 n^ +
· 4 n^ −
· 8 n^ +
· 16 n.
Problem M
We say a real number α is good if there exist nonzero integers m and n such that eαm^ is an integer divisor of 2020n.
(a) Let V denote the set of real numbers which are the sum of two good numbers. Show that V is a Q-vector space under addition.
(b) Calculate dim V and give an example of a basis of V.
Solution
In general, if θm^ is an integer divisor of 2020n^ for some n, then θ = 2x 5 y 101 z^ for some rational numbers x, y, z, with the same sign. This gives a characterization of good numbers. Hence V consists of those β such that eβ^ = 2x 5 y 101 z^ for rational numbers x, y, z. This immediately implies that V is closed under addition and rational multiplication, and so it is indeed a vector space, spanned by log 2, log 5, log 101. By the fundamental theorem of arithmetic it follows readily that these numbers are Q-linearly independent, so they actually form a basis, and in particular dim V = 3.
and so one is nonnegative if and only if the other is. Part (c) is false. One counterexample is to consider a tournament on 16 vertices which has 1 vertex of indegree 5, 10 vertices of indegree 7, and 5 vertices of indegree 9. (One can verify, say by Landau’s theorem, that a tournament with this degree sequence actually exists). In that case we have
A 4 = B 4 = 57440 A 5 = 466440 B 5 = 466560
so this exhibits the desired counterexample.
Problem M
A particle is initially on the number line at a position of 0. Every second, if it is at position x, it chooses a real number t ∈ [− 1 , 1] uniformly and at random, and moves from x to x + t. Find the expected value of the number of seconds it takes for the particle to exit the interval (− 1 , 1). Possible hint: for each 0 < x < 1, let E(x) denote the expected value of the amount of time until the particle exits the interval. You may assume without proof that E(x) is a well-defined and analytic function on the interval (0, 1).
Solution
We let E(x) be the expected value of the time until exiting if the particle starting from a position of x, where 0 < x < 1. For x ≥ 0, obviously E(x) = E(−x) by symmetry. Apparently, E(x) = 1 +
∫ (^) x+
x− 1
E(y) dy.
Let’s define the constant C = 2 +
0 E(y)^ dy.^ Then for 0^ ≤^ x <^ 1, we recover the statement
2 E(x) = C +
∫ (^1) −x
0
E(y) dy.
Actually, note that by setting x = 0 we get 2E(0) = 2C − 2 or E(0) = C − 1. Differentiate once: 2 E′(x) = −E(1 − x) (1)
Differentiate again:
2 E′′(x) = E′(1 − x) = −
E(x) (2)
Consequently, from E′′^ = −E/4 (by (2)) we conclude
E(x) = a sin
x 2
x 2
for some constants a and b, valid for 0 < x < 1. Returning to (1) we should have, for all 0 < x < 1, the identity
a cos
x 2 − b sin
x 2 = −a sin
1 − x 2 − b cos
1 − x 2 = −a
[
sin
cos x 2 − cos
sin x 2
]
− b
[
cos
cos x 2
sin x 2
]
−a sin
− b cos
cos x 2
a cos
− b sin
sin x 2
This can only hold for all 0 < x < 1 if we have
a = −a sin
− b cos
−b = a cos
− b sin
which both imply a b
cos (^12) 1 + sin (^12)
1 − sin (^12) cos (^12)
Problem M
Suppose G is a finite group and ϕ : G → G a homomorphism. Denote by 0 ≤ k ≤ 1 the fraction of elements g ∈ G which satisfy
ϕ(g) = g^2.
(a) Give an example where k = 0.03.
(b) If k 6 = 1, how large can you get k to be?
Solution
This problem is based off an infamous exercise in Herstein’s Topics in Algebra in which the condition ϕ(g) = g^2 is instead ϕ(g) = g−^1. The solution is analogous. For (b), the answer is k = 3/4. An example achieving the equality case is to choose G = Q 8 the quaternion group, and let ϕ(i) = ϕ(j) = −1 and ϕ(k) = +1. To show k > 3 /4 can’t work, we define the set
S =
ϕ(g) = g^2
Fix any s ∈ S. Note that if g ∈ S is an element for which gs ∈ S, then
gsgs = ϕ(gs) = ϕ(g)ϕ(s) = ggss =⇒ gs = sg
and so g is in CG(s), the centralizer of s. By principle of inclusion-exclusion, the number of g ∈ S which have this property is greater than 34 |G| + 34 |G| − |G| = 12 |G|, so in other words, |CG(s)| > 12 |G|. Since CG(s) is a subgroup of G though, we need CG(s) = G. In other words, s lies in the center of G. Thus the center of G contains all of S, but since |S| ≥ 34 |G| > 12 |G|, the center coincides with G — that is, G is abelian. But now if g ∈ G is any element, again we can find some s ∈ S such that gs ∈ S, and now we have
gsgs = ϕ(gs) = ϕ(g)ϕ(s) = ϕ(g)ss =⇒ ϕ(g) = g^2
so ϕ is actually the map g 7 → g^2 on the whole group. Hence k = 1, violating the assumption. As for (a), one may take the product of the quaternion group G = Q 8 × Z/ 25 Z, with ϕ : Q 8 × Z/ 25 Z → Q 8 × Z/ 25 Z acting on the first component as in the previous example and trivially on the second component. This gives 34 · 251 = 0.03.
Problem M
A unit regular tetrahedron is a tetrahedron whose edge lengths are all equal to 1. Two unit regular tetrahedrons ABCD and W XY Z lie in Euclidean space. The labelings of ABCD and W XY Z are oppositely oriented.
(a) How small can max(AW, BX, CY, DZ) be?
(b) Generalize from 3 dimensions to n dimensions.
Solution
The answer to (b) is
(n + 1)/
2 (n^ + 1) 2 ⌋^ and hence the answer to (a) is √ 2 /2. This
problem was suggested by Nikolai Beluhov, who previously proposed the special case n = 2 as problem 4 of grade 10 on the 2012 Autumn Mathematical Tournament in Bulgaria. Let A 1... An+1 and B 1... Bn+1 be the two unit regular n-simplices with opposite orientations. We are going to use two lemmas and one well-known theorem.
Lemma M6.1. Let f : Rn^ → Rn^ be any isometry which flips orientation. Let P be any point, Q = f (P ), and let M denote the midpoint of P Q. As P varies, the locus of M is contained in some (n − 1)-hyperplane.
Proof. Suppose that f maps the orthonormal basis Oe 1 e 2... en onto O′e′ 1 e′ 2... e′ n. Then f is the composition of some linear isometry h with transformation matrix H and the
translation
OO′.
Since h is an isometry, HH>^ = E, and since h flips orientation, det H = −1. We have that H>(E + H) = H>E + H>H = H>^ + E = (E + H)>.
Therefore
− det(E + H) = det H>^ det(E + H) = det
[
H>(E + H)
]
= det(E + H)>^ = det(E + H),
implying that det(E + H) = 0. Now OM~ satisfies
OM^ ~ =^1 2
( OP~ + OQ~ ) =
(E + H) OP~ +
OO~′.
Since det(E + H) = 0, the rank of the transformation matrix E + H is at most n − 1, and so all points M lie in some (n − 1)-hyperplane when P varies, as needed.
Lemma M6.2. We say that two faces of an n-simplex are complementary when their vertex sets form a partitioning of the vertex set of the entire simplex. The shortest distance between two complementary faces of A 1 A 2... An+1 equals
d =
n + 1 ⌊ (^1) 2 (n^ + 1)^2
and is attained exactly when one face is a b(n + 1)/ 2 c-face and the other one is a d(n + 1)/ 2 e-face.
Problem M
Let G be a finite simple graph with n vertices. Say that two Hamiltonian paths P 1 and P 2 of G are neighbors if they have exactly n − 2 edges in common; also say a Hamiltonian path P and a Hamiltonian cycle C of G are neighbors if every edge of P is also an edge of C. Finally, we say that two Hamiltonian cycles C 1 and C 2 of G are equivalent if there exist some number of Hamiltonian paths P 1 , P 2 ,... , Pk of G such that every pair of consecutive terms in the sequence C 1 , P 1 , P 2 ,... , Pk, C 2 are neighbors.
(a) Give an example of a graph G with at least two inequivalent Hamiltonian cycles.
(b) Give an example of a graph G with at least 2020 inequivalent Hamiltonian cycles or prove that no such graph exists.
Solution
This problem is exercise 79 in the current draft of section 7.2.2.4, Hamiltonian Paths and Cycles, of Donald Knuth’s book The Art of Computer Programming. (See https: //cs.stanford.edu/~knuth/fasc8a.ps.gz for the most recent version.) We start with the solution to (a). Let H be the graph whose vertex set is V (H) = Z/ 12 Z, the integers modulo 12. Draw an edge between x and x + 1 for all x ∈ V (H), as well as an edge between y and y + 5 for y ∈ { 0 , 3 , 6 , 9 }, for a total of 16 edges. This gives the graph H below.
This graph has exactly two Hamiltonian cycles, C 1 = 0—1—· · · —11—0 and C 2 = 0— 1—2—9—10—11—6—7—8—3—4—5—0. It is straightforward to check by hand that they are not equivalent. (We need to examine only a small number of cases because the graph is highly symmetric.)
We now proceed to part (b). For the problem, we construct graph G as follows. Let k be any positive integer such that 2k^ ≥ 2020.
- First we take k copies of H, denoted H 0 , H 1 ,... , Hk− 1 , where for all i the vertices of graph Hi are vi,j with 0 ≤ j ≤ 11, so that vertex vi,j of Hi corresponds to vertex j of H. (Thus vertices vi,j′^ and vi,j′′^ of Hi are joined by an edge if and only if vertices j′^ and j′′^ of H are joined by an edge.)
- Create k additional vertices u 0 , u 1 ,... , uk− 1.
- For all i we join vertex ui to vertices vi, 0 and v(i+1 mod k), 1.
This completes our description of graph G. It has a total of 13k vertices and 18k edges.
We claim that G has exactly 2k^ Hamiltonian cycles, and describe them. Every Hamil- tonian cycle of G must contain all edges of G that are incident with a vertex of the form ui; we call these edges of G special. Consequently, every Hamiltonian cycle of G can be obtained as follows:
- Inside each copy Hi of H, we place a copy of either C 1 or C 2.
- Then we delete all edges of the form vi, 0 —vi, 1 from these copies, and we replace them with the 2k special edges of G so as to connect everything up.
Therefore, G has exactly 2k^ ≥ 2020 Hamiltonian cycles. On the other hand, every Hamiltonian path of G can be obtained in one of the following ways:
(i) By omitting one special edge from a Hamiltonian cycle of G.
(ii) By omitting one special edge from a Hamiltonian cycle of G, and then replacing the copy of C 1 or C 2 in the copy Hi of H incident with that special edge by any Hamiltonian path of Hi one of whose endpoints is also a vertex of the other special edge incident with Hi.
(iii) In the exact same way as the Hamiltonian cycles of G, except that in one copy Hi of H we must take a Hamiltonian path of Hi that contains edge vi, 0 —vi, 1 , instead of a copy of either C 1 or C 2.
Since Hamiltonian cycles C 1 and C 2 are not equivalent in H, from this description of the Hamiltonian paths of G we derive that no two Hamiltonian cycles of G are equivalent, either. This concludes the solution.
