I submitted solutions for 20 problems of Nieuw Archief voor Wiskunde:

2005-2/C, 2006-4/AB, 2007-1/AB, 2007-2/C, 2007-4/A, 2008-1/A, 2011-3/A, 2014-1/C, 2014-3/ABC, 2014-4/A, 2015-2/A,
2015-3/ABC, 2015-4/A, 2016-1/C, 2016-2/(A)B(C), 2017-1/AC.

Most of them have already been recognized by NAW.

I received book tokens for 2011-3/A, 2014-3/C, 2015-3/C and 2016-2/C.

**Problem A sept 2011**

*Fix a point P in the interior of a face of a regular tetrahedron Δ. Show that Δ can be partitioned into four congruent convex polyhedra such that P is a vertex of one of them.*

__Solution:__

Let A,B,C,D be the vertices of Δ and suppose P is an arbitrary point in the interior of face ABC.

P is determined by its distances λ, μ, ν to A,B,C resp.

Let us fix three more points so that on each face we have one of the four fixed points and each of A,B,C,D is at distance λ from one of them, at distance μ from another one, and at distance ν from
a third one:

Let T_{ABC} be the point on face ABC at distances λ, μ, ν to A,B,C resp. (So T_{ABC} = P.)

Let T_{BCD} be the point on face BCD at distances λ, μ, ν to C,D,B resp.

Let T_{CDA} be the point on face CDA at distances λ, μ, ν to D,C,A resp.

Let T_{DAB} be the point on face DAB at distances λ, μ, ν to B,A,D resp.

So δ := T_{ABC}T_{BCD}T_{CDA}T_{DAB} is also a regular tetrahedron. Let O be its center.

Now we will allot points of Δ to each X ∈ {A,B,C,D} in such a way that the points allotted to X form a convex polyhedron, and such that the four polyhedra are congruent and form a partition
of Δ.

The allotment will first be done in two steps:

Step 1)

The points in the hexahedron AT_{ABC}T_{CDA}T_{DAB}O belong to A.

The points in the hexahedron BT_{ABC}T_{BCD}T_{DAB}O belong to B.

The points in the hexahedron CT_{ABC}T_{BCD}T_{CDA}O belong to C.

The points in the hexahedron DT_{BCD}T_{CDA}T_{DAB}O belong to D.

Step 2)

To allot the remaining points in the interior of Δ while preserving convexity and congruence, we extend (to the exterior of δ)
the six planes through O and a side of δ that bisect the angles between two faces of δ.

Through each vertex of δ there go three of these bisectrix planes.

Now for each X ∈ {A,B,C,D}, allot to X all remaining points y ∈ Δ such that, for each plane through O and a side of the nearest face of δ,
X and y are at the same side of that plane.

Of course, we can sum it all up, including both steps 1 and 2, as follows:

For each X ∈ {A,B,C,D}, allot to X all points y ∈ Δ such that, for each plane through O and a side of the nearest face of δ,
X and y are at the same side of that plane.

**Problem 2A 2013 revised (to make it a bit simpler)**

*We have two hourglasses, A for a seconds and B for b seconds, where a and b are relatively prime positive integers.
Let t be an integer with t ≥ ab. Show that A and B can be used to identify the time t if the upper bulbs are empty at the start.*

__Solution:__

We will find non-negative integers p and q such that pa + qb = t, which is clearly sufficient for our purpose, as follows;

Since a and b are coprime, there exist positive integers m and n such that mb - na = 1 (or mb - na = -1, then reverse the roles of a and b).

So for any positive integer s we have (tm-sa)b + (sb-tn)a = t.

If we can find such s within [tn/b,tm/a] = [tn/b, tn/b + t/ab], we can take p = tm-sa, q = sb-tn, and then we are finished.

Now if t ≥ ab, then this interval has length ≥ 1, so then we can find in it this positive integer s.

**Problem 4A 2014 **

*Let k be an integer, at least 4. Determine the variance of the greatest common divisor of k positive integers.
Here we mean the limit, as n → ∞, of the variance of the greatest common divisor of k integers in {1,2,...,n} with respect to the uniform distribution on {1,2,...,n} ^{k}.*

__Solution:__

We find the probability that d is gcd is (1/d^{k})*Π_{p prime}(1 - 1/p^{k}) = 1/(d^{k}*ζ(k)).

So the expectation value of the gcd is Σ_{d=1}^{∞} d/(d^{k}*ζ(k)) = ζ(k-1)/ζ(k).

So the variance of the gcd is Σ_{d=1}^{∞} (d - ζ(k-1)/ζ(k))^{2}/(d^{k}*ζ(k)) =
(ζ(k)*ζ(k-2) - ζ(k-1)^{2})/ζ(k)^{2}.

(The formula for the probability means: that the k integers are divisible by d and, after division by d, for each prime p, not all of them divisible by p.)

**Problem 1C 2016**

* Determine all two-sided infinite sequences of positive integers in which each number is the Euler-φ of the next.*

__Solution__

If n has prime factorization p_{1}^{a1}p_{2}^{a2}...p_{k}^{ak} then
φ(n) = p_{1}^{a1-1}p_{2}^{a2-1}...p_{k}^{ak-1}(p_{1}-1)(p_{2}-1)...(p_{k}-1).

Since for n larger than 1 we find φ(n) smaller than n, such a sequence must begin with infinitely many 1's.

We may terminate these 1's at the left hand side of the sequence by replacing the last 1 by 2, which is the only other number whose φ is 1.

If we do that, then in the next position after 2 we can put only 2^{2} or 2*3. We can't put 3, because there would be no option for the next position after the 3.

Now if at some position we have 2^{s}, then in the next position we can only put 2^{s+1} or 2^{s}3.
For example, we can't put 2^{s-1}5 or 2^{s-3}17, for, going further to the right, we would soon run out of factors 2.

And if at some position we have 2^{s}3^{t}, then in the next position we can only put 2^{s}3^{t+1}. For example, we can't put there 2^{s-1}3^{t}7,
because we would again run out of options for lack of factors 2 after a few steps to the right.

So this sums up all possible sequences.

**Problem 2B 2016**: Suppose there are N ≥ 2 players, labeled 1,2,..,N, and each of them holds precisely m ≥ 1 coins of value 1, m coins of (integer) value n ≥ 2, m coins of value n^{2}, etc.

A transaction from player i to player j consists of player i giving a finite number of his coins to player j.

We say an N-tuple (a_{1},a_{2}, ... ,a_{N}) of integers is (m,n)-payable if Σ_{i=1,2,..,N} a_{i} = 0 and after a finite number of transactions the
i-th player has received (in value) a_{i} more than he has given away.

Show that for every N-tuple (a_{1},a_{2}, ... ,a_{N}) with Σ_{i=1,2,..,N} a_{i} = 0 to be (m,n)-payable, it is necessary and sufficient that
m > n - n/N - 1.

**Solution**

The inequality can be rewritten as (n-1)(N-1) ≤ mN.

This is the necessary and sufficient condition that for each k the mN coins can be redistributed so that all but one players at the end have up to n-1 coins of worth n^{k} for each k,
thus covering in the n-ary system all values of possession.

(The player who paid the most holds the rest of the total value.)