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306

NAW 5/17 nr. 4 december 2016 Problemen

Pr obl em en

Redactie:

Gabriele Dalla Torre Christophe Debry Jinbi Jin Marco Streng Wouter Zomervrucht Problemenrubriek NAW Mathematisch Instituut Universiteit Leiden Postbus 9512 2300 RA Leiden

problems@nieuwarchief.nl www.nieuwarchief.nl/problems

| Problem Section

This Problem Section is open to everyone; everybody is encouraged to send in solutions and propose problems. Group contributions are welcome.

For each problem, the most elegant correct solution will be rewarded with a book token worth € 20. At times there will be a Star Problem, to which the proposer does not know any solution. For the first correct solution sent in within one year there is a prize of € 100.

When proposing a problem, please either include a complete solution or indicate that it is intended as a Star Problem.

Please send your submission by e-mail (LaTeX is preferred), including your name and ad- dress to problems@nieuwarchief.nl.

The deadline for solutions to the problems in this edition is 1 March 2017.

Problem A (folklore)

For a finite sequence s=( ,s₁f, )s_n of positive integers, denote by ( )p s the number of ways to write s as a sum s=

/

_iⁿ₌₁a e_{i i}+

/

ⁿ_j=^-₁¹b e_j( _j+e_j+1) with all a_i and b_j non-negative. Here e_i denotes the sequence of which the i-th term is 1 and of which all the other terms are 0.

Show that there exists an integer B> such that for any product F of (positive) Fibonacci 1 numbers, there exists a finite sequence s=( ,s₁f, )s_n with all s_i!{ , ,1 2f, }B such that

( ) p s = .F

Problem B (folklore)

Let , be a prime number. For any group homomorphism :f A"B between abelian groups and for any integer n$0, denote by f_n the induced homomorphism /A ,ⁿA"B/,ⁿB. Let ( )k_{n n 0}3₌ and ( )c_{n n 0}³₌ be sequences of integers.

Show that there exist integers , ,N a b$0 and a group homomorphism :( /f Z ,^NZ)^a"( /Z ,^NZ)^b such that for all n$0 we have # ker f_n=,^kn and # coker f_n=,^cn if and only if k₀=c₀= 0 and the sequences (k_n+1-k_{n n})³₌₀ and (c_n+1-c_{n n})³₌₀ are non-negative, non-increasing, eventually zero, and there is a constant C such that for all n such that k_n+1-k_n and c_n+1-c_n are not both zero, their difference is C.

(Recall that the cokernel coker f of a group homomorphism :f A"B between abelian groups is the quotient of B by the image of f.)

Problem C (folklore)

Let R be the polynomial ring over Z with variables x_i, y_i, z_i for all i!Z. Let S be the polynomial ring over Z with variables t_i for all i!Z. Let : Rx "R be the isomorphism of rings given by x_i7x_{i 1+} , y_i7y_{i 1+} and z_i7z_{i 1+} .

Consider the morphism :f R"S of rings given by x_i7t_i-1t t_{i i+1}, y_i7t_i³ and z_i7t_i². Does there exist a finite number of elements , ,r₁fr_n!R such that the kernel I of f is generated as an ideal in R by {xⁱr i_j: !Z,j=1,f, }n?

Edition 2016-2 We received solutions from Hendrik Reuvers, Pieter de Groen en Toshihiro Shimizu.

Problem 2016-2/A (folklore)

Denote for all positive rational numbers x by f x^ h the minimum number of 1’s needed in a formula for x involving only ones, addition, subtraction, multiplication, division and parentheses. For example, ( )f 1 = , and ( )1 f ₃¹ = , as 4 ₃¹=_{1 1 1+ +}¹ and as no such formula exists with at most three 1’s. Note that ( )f 11 !2 (concatenation of ones is not allowed).

Moreover, denote for all positive rational numbers x by ( )h x₂ the number log₂( )p +log₂( )q, where log₂ denotes the base-2 logarithm, and where p, q are positive integers such that x= and _q^p gcd p q( , )= .1

Show that for all x, we have

( ) ( ).

f x >¹₂h x2

Problemen NAW 5/17 nr. 4 december 2016

307

Opl ossin gen

| Solutions _Solution We received a solution from Toshihiro Shimizu, to whom the book token is award-

ed, and the solution of whom the following is based on.

Let x be a positive rational number, and let p, q be positive integers such that x qp

= and ( , )

gcd p q = . We show by induction on ( )1 f x that ,p q#2^{f x( )-1}; from this it will follow that

( ) ( ) log log ( )

f x f x p q h x

2 >2 -2$ 2 + 2 = 2 , as desired.

First note that x= is the only rational number with ( )1 f x = , and in this case p1 = = = q 1 2^{f x( )-1}. So suppose that n is a positive integer and that for any x with ( )f x #n written as qp

with p, q positive integers with gcd p q( , )= , we have ,1 p q#2^{f x( )-1}. Suppose that ( )

f x = + . Then x can be written in one of the forms xn 1 1+ , xx2 1- , x xx2 1$ 2, xx 2 1with x1, x2 positive rational numbers with ( )f x1 +f x( )2 =f x( )= + (so that ( ), ( )n 1 f x1 f x2 #n).

Write xi qp i

= iwhere pi and qi are positive integers such that gcd p q( , )i i = , for 1 i=1 2, . If x=x1+ or xx2 =x1- , then we find that xx2 p qq qp q

1 2 1 2 2 1

= ^! , so ,p q#2^{1+f x( )1- +1 f x( )2-1}=2ⁿ 2^{f x( ) 1}

= ^- . If x=x x1$ 2, then we find that x q qp p 1 2

= 1 2, so ,p q#2^{f x( )1- +1 f x( )2-1}=2^n-1<2^{f x( )-1}. The same argument shows that if x xx

2

= 1, that then ,p q<2^{f x( )-1}. Hence we always have ,

p q#2^{f x( )-1}, and we are done.

Problem 2016-2/B (folklore)

Suppose that there are N$2 players, labeled , , ,N1 2 f , and that each of them holds precisely m$1 coins of value 1, m coins of (integer) value n$2, m coins of value n², et cetera. A transaction from player i to player j consists of player i giving a finite number of his coins to player j. We say that an N-tuple ( , , ,a a1 2faN) of integers is ( , )m n -payable if i ai 0

N

1 =

/

= and after a finite number of transactions, the i-th player has received (in value) ai more than he has given away.

Show that for every N-tuple ( , , ,a a1 2faN) with

/

_i^N₌₁a_i=0 to be ( , )m n -payable, it is nec- essary and sufficient that m>n-_Nⁿ- .1

Solution We received solutions from Pieter de Groen and Toshihiro Shimizu. The book to- ken goes to Pieter de Groen. Both solutions shared the same idea, the following solution is based on that of Toshihiro Shimizu.

First observe that if an N-tuple ( , , , )a a1 2fan is ( , )m n -payable, the number of coins of value 1 that player i receives is ai modulo n, since coins of higher value are of value divisible by n.

Also observe that m>n-_Nⁿ - is equivalent to 1 mN$(n-1)(N- .1)

Note that if m#n-_Nⁿ - (so m1 < ), or equivalently, n mN<(n-1)(N- , then the tuple1)

, , , , ( )( )

n m- -1n m- -1fn m- -1- N-1 n m- -1

^ h

is not ( , )m n -payable; every player up to player N 1- has to either receive at least n m 1- - coins of value 1, or give away at least m 1+ coins of value 1. The latter is clearly impos- sible. This means that the last player must give away at least (N-1)(n m- - coins 1) of value 1. However, (N-1)(n m- -1)=(n-1)(N-1)-m N( -1 >) m, so this is also impossible. Therefore the condition m>n-_Nⁿ - is necessary.1

Next, we show that the condition m>n-_Nⁿ - is sufficient. Let ( ,1 a₁f,a_N) be any N-tuple of integers with

/

_ia_i=0. Assume that m>n-_Nⁿ - . We show that ( ,1 a₁f,a_N) is ( , )m n - payable, by induction on max a_i| _i|.

If max a_i| _i|#1, then all a_i lie in {-1 0 1, , }, and as the sum of the a_i’s is zero, the tuple is ( , )m n -payable (by having every player with negative a_i pay one coin of value 1, and every player with positive a_i receive one coin of value 1). So assume that A> , and that all 1 tuples ( , ,a₁fa_N) with zero sum and max a_i| _i|<A are ( , )m n -payable.

Let ( , ,a₁fa_N) be a tuple with max a_i| _i|= . Let rA _i denote the remainder of a_i on division by n. We assume without loss of generality that n>r₁$r₂$g$r_N$0. Note that

/

_{i i}r^is a multiple of n, say

/

_ir_i=kn. We show that n r- ₁,f,n r- _k#m, so that player i can pay n r- coins of value 1 for _i i=1 2 f, , ,k.

Note that

( ) ( ) ( ) ( ) .

kn r_i r r r r N k 1 r k 1 n

i 1 g k 1 k g N # k

=

/

= + + _- + + + - + + -

Therefore r_k$ _{N k- +}ⁿ ₁ $ _Nⁿ, so n r- _k= -n _Nⁿ <m 1+ , from which we deduce that n r- _k m

# , and therefore also that n r- _i#m for all i=1 2 f, , ,k. So by having, for i=1 2 f, , ,k,

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NAW 5/17 nr. 4 december 2016 Problemen

Opl ossin gen

| Solutions player i pay n r_i- , and for all other i, player i receive r_i (which is possible by the

above and since

/

_ir_i=kn), we see that the tuple ( , ,a₁fa_N) is ( , )m n -payable if (a₁+ -n r₁,f,a_k+ -n r a_k, _k+1-r_k+1,f,a_N-r_N) is ,_^m n_h-payable using only coins of val-

ue n or higher, and therefore if the tuple

’ ’

( ,a1 ,aN) n1(a n r), , (n ak n rk), (n ak rk ), , (n aN rN)

1 1 1 1

1 1 1

f =_ + - f + - ₊ - ₊ f - i

is ( , )m n -payable.

We finish the induction by showing that max ai| ’i|<A, so that by the induction hypoth- esis, the tuple ’( ,a1f,a’N)is indeed ( , )m n -payable. Note that a’i=` j or ’an_i ai=8 B, so an_i

’

|ai|#` j. Since <sup>| |</sup>an<sub>i</sub> ` j^{| |a}_nⁱ #|ai| with equality if and only if | |ai = or | |0 ai = , it follows 1 that ’|ai| |< ai| for all i with | |ai $2. Since A$2, it follows that max ai| ’i|<A, and we are done.

Problem 2016-2/C (proposed by Wouter Zomervrucht)

For each integer n$1 let cn be the largest real number such that for any finite set of vectors X1Rⁿ with

/

_{v X!} | |v $1 there exists a subset Y3X with |

/

_{v Y!} v|$cn. Prove the recurrence relation

, .

c1 12 cn nc 1 21

= + = _r n

Solution We received solutions from Hendrik Reuvers and Toshihiro Shimizu. The book token is awarded to Hendrik Reuvers. Both solutions shared the same idea, the following solution is based on the one sent in by the proposer.

For n$0, let Dⁿ1Rⁿ be the closed unit ball and Sⁿ1R^{n 1+} the unit sphere. Denote by vn the volume of Dⁿ and by sn the surface area of Sⁿ. We will show that cn=vn-1/sn-1, then we are done by the relations v0= , s1 0= , v2 n n n1s

= -1, and sn=2rvn 1- .

First we make a computation. Let n$1 and write V+={x!Rⁿ:x_n$0}. For r> we let 0 rDⁿ be the closed radius r ball, then one has

( ) ,

x dx x r x v dx n

v 1r

n rD V

n n n n

r

n n

2 2

1 0

1 1

n

n21

= - = +

+

- - +

+

# #

-

so

. x dx_n drd x dx v

S V

n

rD V r

n 1

1

n 1 n

= =

+ + =

-

- + +

R

T SSSS SSSS

V

X WWWW WWWW

# #

Now we turn to the problem. Take n$1 and any collection X1Rⁿ with

/

_{v X!} | |v $1^{. Let} Y3X be a subset for which the subsum w=

/

_{v Y!} v has maximal norm. Then Y must contain all v!X with v w$ >0, and no v!X with v w$ <0. In fact, | |w =m w( / | |)w , where for any x!S^{n 1-} we define

( ) .

m x v x

with

v X v x 0

= $

$

!

/

$

It follows that | |w =max_{x S}! ^{n 1-} m x( ). For v!Rⁿ set V_v={x!Rⁿ:v x$ $0}. (For instance, V+=V_{( , , , )0}f_{0 1}.) We compute

( ) | | .

m x dx v x dx v x dx v

S v XS V v X n

S V

n 1

n n

v n

1 1 1

$ $

= =

+ +

! ! -

- - -

+

/ /

# # #

Thus there is x!S^{n 1-} where ( )m x $v_n-1/s_n-1, hence c_n$v_n-1/s_n-1.

Conversely, let X consist of k vectors with lengths /k1 and directions distributed homo- geneously over S^{n 1-} . (There are several ways of doing this; thanks to Toshihiro Shi- mizu for pointing out [1].) As k " 3, the associated function m converges uniformly to a constant function; by the computations above, its constant value is v_n-1/s_n-1. So also

/ c_n#v_n-1 s_n-1. Reference

1 Eric W. Weisstein, Sphere Point Picking, http://mathworld.wolfram.com/SpherePointPicking.html.