Summer Workshop in Mathematics
Solutions to SWIM 2016 Problems
June 13–June 23, 2016
Department of Mathematics, Duke University, Durham NC 27708
Question 1
Question: A group of n people are standing in a circle, numbered consecutively clockwise from 1
to n. Starting with person 2, we remove every other person, proceeding clockwise. For example,
if n = 5, the people are removed in the order 2, 4, 1, 5, 3 and the last person remaining is 3. Let
f (n) denote the last person remaining.
(a) Compute f (n) for n = 2, 3, · · · , 25.
(b) Find a way to compute f (n) for any positive integer n > 1.
Solution: For (a), one can easily obtain the values of f (n) for n = 2, 3, . . . , 25:
n
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
f (n) 1 3 1 3 5 7 1 3 5 7 9 11 13 15 1 3 5 7 9 11 13 15 17 19
For (b), we will prove that if n = 2k + a, where 0 ≤ a < 2k , then f (n) = 2a + 1. Observe that there
are recurrence relations, namely f (2m) = 2f (m) − 1 and f (2m + 1) = 2f (m) + 1 for any m ≥ 2:
For instance, suppose that n = 2m. Then we first remove m people, namely 2, 4, 6, . . . , 2m, so
that there are m people left, namely 1, 3, 5, . . . , 2m − 1. Now, we want to claim that there is a
one to one correspondence between A = {1, 3, . . . , 2m − 1} and B = {1, 2, 3, . . . , m} when we are
removing the people from each set following the rule. As in order, we remove the person from A as
3, 7, 11 so on. Respectively, we remove the person from B as 2, 4, 6 so on. Notice that 3 = 2 · 2 − 1,
7 = 2 · 4 − 1, 11 = 2 · 6 − 1 so on. In other words, if we denote α the number removed from A
and β the corresponding number removed from B, then we obtain the first relation α = 2β − 1.
Therefore, one knows that f (2m) the last person removed from A corresponds to f (m) the last
person removed from B and hence they satisfy f (2m) = 2f (m) − 1. Similarly, one obtains the
second relation f (2m + 1) = 2f (m) + 1.
With these in mind, we now prove that, by induction on k, if n = 2k + a, 0 ≤ a < 2k , then
f (n) = 2a + 1: First if k = 1, then a is either 0 or 1. From the table above, f (2) = 1 = 2 · 0 + 1
and f (3) = 3 = 2 · 1 + 1. Therefore the base step is true. Assume now that the statement is true
for k. In other words, assume that f (2k + a) = 2a + 1. We will need to prove that this is also true
for k + 1.
If a is even, say a = 2a0 , then we have that
f 2k+1 + a = f 2(2k + a0 ) = 2f (2k + a0 ) − 1 = 2(2a0 + 1) − 1 = 2a + 1.
1
2
Here in the second equality, we employ the first recurrence relation for the even number above
and in the third equality, we use the induction hypothesis.
Similarly, if a is odd, say a = 2a0 + 1, then one has that
f 2k+1 + a = f 2(2k + a0 ) + 1 = 2f (2k + a0 ) + 1 = 2(2a0 + 1) + 1 = 2a + 1.
Again, in the second equality, we employ the second recurrence relation for the odd number above
and in the third equality, we use the induction assumption. Therefore we prove that the statement
is true for k + 1 in both cases and so we complete the proof.
Question 2
Question: Start with a set of lattice points in the plane, i.e. pairs (a, b) with a and b both integers.
You can repeatedly produce new points and enlarge the set by applying any of the following rules,
in any order:
(1) If (a, b) is in the set then add the point (a + 1, b + 1) to the set.
(2) If a and b are both even and (a, b) is in the set then add (a/2, b/2) to the set.
(3) If the pair of points (a, b) and (b, c) are in the set then add the point (a, c) to the set.
If your original set is the single point (7, 29), is it possible eventually apply these rules to construct
a set containing the point (3, 1999)?
Solution: Suppose that the point (a, b) is given. Observe that any point (a + k, b + k), k ∈ N,
obtained by applying the rule (1) to (a, b) will have the same coordinate difference as the original
point (a, b), namely b − a. Similarly, if we apply the rule (2) to (a, b), where both a and b are
even, to obtain ( a2 , 2b ), then its coordinate difference will be then (b−a)
. If we apply the rule (3) to
2
(a, b) and the new point (b, c) obtained by applying either the rule (1) or (2), then the coordinate
difference c − a of the new point (a, c) should be a multiple of (b−a)
because c − a = (b − a) + (c − b).
2
Therefore, for any two points (α, β) and (β, γ) obtained by applying either the rule (1) or the
rule (2), one knows that the coordinate difference γ − α for any new point (α, γ) will be again a
because
multiple of (b−a)
2
γ − α = γ − β +β − α
| {z } | {z }
∗
∗∗
.
and ∗ and ∗∗ are both multiples of (b−a)
2
Since the coordinate difference of the original point (7, 29) is 22, so whatever points one obtains
should be a multiple of 11. Clearly, the coordinate difference (3, 1999) is 1996 which is not a
multiple of 11. Therefore, it is not possible to obtain (3, 1999) if one starts from (7, 29).
Question 3
Question: The first quadrant is the set of (x, y) in the plane with x ≥ 0, y ≥ 0, and the unit
circle is the set of (x, y) in the plane such that x2 + y 2 = 1. Let s be any arc of the unit circle
lying entirely in the first quadrant. Let A be the area of the region lying below s and the above
3
x-axis and let B be the area of the region lying to left of s and to the right of the y-axis. Prove
that A + B depends only on the arc length, not on the position of s.
Solution: Compare the following three pictures:
θ
θ
θ
Let s be the given arc whose angle is fixed as θ (in radian) as in the above figure. Let A be the
area of the region shade in red and B be the area of the region shade in green. One might guess,
based on the above figure, that A + B does depend only on θ, not on the position of s. Indeed,
that is the case. Consider the following picture:
Note that A + B = CDEF + F GHI + 2(gEF G), where CDEF and F GHI denote the
area of the rectangle CDEF (red) and F GHI (green), respectively, and let gEF G denote the
area of the region colored in both red and green. If we let 4OEF be the area of the red triangle
OEF , one knows CDEF = 24OEF . Similarly, one has F GHI = 24OF G for the green.
Therefore, we conclude that
A + B =CDEF + F GHI + 2(gEF G) = 24OEF + 24OF G + 2(gEF G)
=2 4OEF + 24OF G + gEF G = 2(θ/2) = θ,
which clearly depends only on θ. Here θ/2 is then the area of the pizza slice gOEG. This
completes the proof.