diff --git a/README.md b/README.md
index a008d00..3756a52 100644
--- a/README.md
+++ b/README.md
@@ -429,15 +429,16 @@ Problem 19
 You are given the following information, but you may prefer to do some
 research for yourself.
 
-     • 1 Jan 1900 was a Monday.
-     • Thirty days has September,
-       April, June and November.
-       All the rest have thirty-one,
-       Saving February alone,
-       Which has twenty-eight, rain or shine.
-       And on leap years, twenty-nine.
-     • A leap year occurs on any year evenly divisible by 4, but not on a
-       century unless it is divisible by 400.
+*   1 Jan 1900 was a Monday.
+
+*   Thirty days has September,  
+    April, June and November.  
+    All the rest have thirty-one,  
+    Saving February alone,  
+    Which has twenty-eight, rain or shine.  
+    And on leap years, twenty-nine.
+
+*   A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
 
 How many Sundays fell on the first of the month during the twentieth
 century (1 Jan 1901 to 31 Dec 2000)?
@@ -870,18 +871,18 @@ Problem 43
    up of each of the digits 0 to 9 in some order, but it also has a rather
    interesting sub-string divisibility property.
 
-   Let d[1] be the 1st digit, d[2] be the 2nd digit, and so on. In this
-   way, we note the following:
+Let d[1] be the 1st digit, d[2] be the 2nd digit, and so on. In this
+way, we note the following:
 
-     • d[2]d[3]d[4]=406 is divisible by 2
-     • d[3]d[4]d[5]=063 is divisible by 3
-     • d[4]d[5]d[6]=635 is divisible by 5
-     • d[5]d[6]d[7]=357 is divisible by 7
-     • d[6]d[7]d[8]=572 is divisible by 11
-     • d[7]d[8]d[9]=728 is divisible by 13
-     • d[8]d[9]d[10]=289 is divisible by 17
+*   $d_2d_3d_4=406$ is divisible by 2
+*   $d_3d_4d_5=063$ is divisible by 3
+*   $d_4d_5d_6=635$ is divisible by 5
+*   $d_5d_6d_7=357$ is divisible by 7
+*   $d_6d_7d_8=572$ is divisible by 11
+*   $d_7d_8d_9=728$ is divisible by 13
+*   $d_8d_9d_10=289$ is divisible by 17
 
-   Find the sum of all 0 to 9 pandigital numbers with this property.
+Find the sum of all 0 to 9 pandigital numbers with this property.
 
    
    Answer: 115253b7721af0fdff25cd391dfc70cf
@@ -1893,21 +1894,21 @@ will remain on the CC/CH square.
 
 Community Chest (2/16 cards):
 
-1. Advance to GO
-2. Go to JAIL
+1.  Advance to GO
+2.  Go to JAIL
 
 Chance (10/16 cards):
 
-1. Advance to GO
-2. Go to JAIL
-3. Go to C1
-4. Go to E3
-5. Go to H2
-6. Go to R1
-7. Go to next R (railway company)
-8. Go to next R
-9. Go to next U (utility company)
-10. Go back 3 squares.
+1.  Advance to GO
+2.  Go to JAIL
+3.  Go to C1
+4.  Go to E3
+5.  Go to H2
+6.  Go to R1
+7.  Go to next R (railway company)
+8.  Go to next R
+9.  Go to next U (utility company)
+10. Go back 3 squares.
 
 The heart of this problem concerns the likelihood of visiting a particular
 square. That is, the probability of finishing at that square after a roll.
@@ -2401,8 +2402,8 @@ Let S(A) represent the sum of elements in set A of size n. We shall call
 it a special sum set if for any two non-empty disjoint subsets, B and C,
 the following properties are true:
 
-i. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
-ii. If B contains more elements than C then S(B) > S(C).
+1.  S(B) ≠ S(C); that is, sums of subsets cannot be equal.
+2.  If B contains more elements than C then S(B) > S(C).
 
 If S(A) is minimised for a given n, we shall call it an optimum special
 sum set. The first five optimum special sum sets are given below.
@@ -3099,23 +3100,23 @@ Problem 127
    The radical of n, rad(n), is the product of distinct prime factors of n.
    For example, 504 = 2^3 × 3^2 × 7, so rad(504) = 2 × 3 × 7 = 42.
 
-   We shall define the triplet of positive integers (a, b, c) to be an
-   abc-hit if:
+We shall define the triplet of positive integers (a, b, c) to be an
+abc-hit if:
 
-    1. GCD(a, b) = GCD(a, c) = GCD(b, c) = 1
-    2. a < b
-    3. a + b = c
-    4. rad(abc) < c
+1.  GCD(a, b) = GCD(a, c) = GCD(b, c) = 1
+2.  a < b
+3.  a + b = c
+4.  rad(abc) < c
 
-   For example, (5, 27, 32) is an abc-hit, because:
+For example, (5, 27, 32) is an abc-hit, because:
 
-    1. GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1
-    2. 5 < 27
-    3. 5 + 27 = 32
-    4. rad(4320) = 30 < 32
+1.  GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1
+2.  5 < 27
+3.  5 + 27 = 32
+4.  rad(4320) = 30 < 32
 
-   It turns out that abc-hits are quite rare and there are only thirty-one
-   abc-hits for c < 1000, with ∑c = 12523.
+It turns out that abc-hits are quite rare and there are only thirty-one
+abc-hits for c < 1000, with ∑c = 12523.
 
    Find ∑c for c < 120000.
 
@@ -5763,19 +5764,20 @@ Problem 236
    Although the suppliers try very hard to ship their goods in perfect
    condition, there is inevitably some spoilage - i.e. products gone bad.
 
-   The suppliers compare their performance using two types of statistic:
+The suppliers compare their performance using two types of statistic:
 
-     • The five per-product spoilage rates for each supplier are equal to the
-       number of products gone bad divided by the number of products
-       supplied, for each of the five products in turn.
-     • The overall spoilage rate for each supplier is equal to the total
-       number of products gone bad divided by the total number of products
-       provided by that supplier.
+*   The five per-product spoilage rates for each supplier are equal to the
+    number of products gone bad divided by the number of products
+    supplied, for each of the five products in turn.
 
-   To their surprise, the suppliers found that each of the five per-product
-   spoilage rates was worse (higher) for 'B' than for 'A' by the same factor
-   (ratio of spoilage rates), m>1; and yet, paradoxically, the overall
-   spoilage rate was worse for 'A' than for 'B', also by a factor of m.
+*   The overall spoilage rate for each supplier is equal to the total
+    number of products gone bad divided by the total number of products
+    provided by that supplier.
+
+To their surprise, the suppliers found that each of the five per-product
+spoilage rates was worse (higher) for 'B' than for 'A' by the same factor
+(ratio of spoilage rates), m>1; and yet, paradoxically, the overall
+spoilage rate was worse for 'A' than for 'B', also by a factor of m.
 
    There are thirty-five m>1 for which this surprising result could have
    occurred, the smallest of which is 1476/1475.
@@ -6359,12 +6361,12 @@ Problem 258
 ===========
 
 
-   A sequence is defined as:
+A sequence is defined as:
 
-     • g[k] = 1, for 0 ≤ k ≤ 1999
-     • g[k] = g[k-2000] + g[k-1999], for k ≥ 2000.
+*   $g_k = 1$, for $0 ≤ k ≤ 1999$
+*   $g_k = g_{k-2000} + g_{k-1999}$, for $k ≥ 2000$.
 
-   Find g[k] mod 20092010 for k = 10^18.
+Find g[k] mod 20092010 for k = 10^18.
 
    
    Answer: 18eca0138f3acbde20dcc24ed06627ea
@@ -6374,20 +6376,17 @@ Problem 259
 ===========
 
 
-   A positive integer will be called reachable if it can result from an
-   arithmetic expression obeying the following rules:
+A positive integer will be called reachable if it can result from an
+arithmetic expression obeying the following rules:
 
-     • Uses the digits 1 through 9, in that order and exactly once each.
-     • Any successive digits can be concatenated (for example, using the
-       digits 2, 3 and 4 we obtain the number 234).
-     • Only the four usual binary arithmetic operations (addition,
-       subtraction, multiplication and division) are allowed.
-     • Each operation can be used any number of times, or not at all.
-     • Unary minus is not allowed.
-     • Any number of (possibly nested) parentheses may be used to define the
-       order of operations.
+*   Uses the digits 1 through 9, in that order and exactly once each.
+*   Any successive digits can be concatenated (for example, using the digits 2, 3 and 4 we obtain the number 234).
+*   Only the four usual binary arithmetic operations (addition, subtraction, multiplication and division) are allowed.
+*   Each operation can be used any number of times, or not at all.
+*   Unary minus is not allowed.
+*   Any number of (possibly nested) parentheses may be used to define the order of operations.
 
-   For example, 42 is reachable, since (1/23) * ((4*5)-6) * (78-9) = 42.
+For example, 42 is reachable, since (1/23) * ((4*5)-6) * (78-9) = 42.
 
    What is the sum of all positive reachable integers?
 
@@ -6404,13 +6403,13 @@ Problem 260
    if she takes stones from more than one pile, she must remove the same
    number of stones from each of the selected piles.
 
-   In other words, the player chooses some N>0 and removes:
+In other words, the player chooses some N>0 and removes:
 
-     • N stones from any single pile; or
-     • N stones from each of any two piles (2N total); or
-     • N stones from each of the three piles (3N total).
+*   N stones from any single pile; or
+*   N stones from each of any two piles (2N total); or
+*   N stones from each of the three piles (3N total).
 
-   The player taking the last stone(s) wins the game.
+The player taking the last stone(s) wins the game.
 
    A winning configuration is one where the first player can force a win.
    For example, (0,0,13), (0,11,11) and (5,5,5) are winning configurations
@@ -6443,14 +6442,14 @@ Problem 261
 
                  (k-m)^2 + ... + k^2 = (n+1)^2 + ... + (n+m)^2.
 
-   Some small square-pivots are
+Some small square-pivots are
 
-     • 4: 3^2 + 4^2 = 5^2
-     • 21: 20^2 + 21^2 = 29^2
-     • 24: 21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2
-     • 110: 108^2 + 109^2 + 110^2 = 133^2 + 134^2
+*   $\boldsymbol{4}: 3^2 + \boldsymbol{4}^2 = 5^2$
+*   $\boldsymbol{21}: 20^2 + \boldsymbol{21}^2 = 29^2$
+*   $\boldsymbol{24}: 21^2 + 22^2 + 23^2 + \boldsymbol{24}^2 = 25^2 + 26^2 + 27^2$
+*   $\boldsymbol{110}: 108^2 + 109^2 + \boldsymbol{110}^2 = 133^2 + 134^2$
 
-   Find the sum of all distinct square-pivots ≤ 10^10.
+Find the sum of all distinct square-pivots ≤ 10^10.
 
    
    Answer: d45ddf64010ed143228a6a6b84837de9
@@ -6507,12 +6506,12 @@ Problem 263
    prime pairs, such that the second member of each pair is the first member
    of the next pair.
 
-   We shall call a number n such that :
+We shall call a number n such that :
 
-     • (n-9, n-3), (n-3,n+3), (n+3, n+9) form a triple-pair, and
-     • the numbers n-8, n-4, n, n+4 and n+8 are all practical,
+*   $(n-9, n-3), (n-3,n+3), (n+3, n+9)$ form a triple-pair, and
+*   the numbers $n-8$, $n-4$, $n$, $n+4$ and $n+8$ are all practical,
 
-   an engineers’ paradise.
+an engineers’ paradise.
 
    Find the sum of the first four engineers’ paradises.
 
@@ -6525,9 +6524,9 @@ Problem 264
 
 Consider all the triangles having:
 
-* All their vertices on lattice points.
-* Circumcentre at the origin O.
-* Orthocentre at the point H(5, 0).
+*   All their vertices on lattice points.
+*   Circumcentre at the origin O.
+*   Orthocentre at the point H(5, 0).
 
 There are nine such triangles having a perimeter ≤ 50.
 Listed and shown in ascending order of their perimeter, they are:
@@ -7161,16 +7160,17 @@ Problem 292
 ===========
 
 
-   We shall define a pythagorean polygon to be a convex polygon with the
-   following properties:
+We shall define a pythagorean polygon to be a convex polygon with the
+following properties:
+
+*   there are at least three vertices,
+*   no three vertices are aligned,
+*   each vertex has integer coordinates,
+*   each edge has integer length.
 
-     • there are at least three vertices,
-     • no three vertices are aligned,
-     • each vertex has integer coordinates,
-     • each edge has integer length.
+For a given integer n, define P(n) as the number of distinct pythagorean
+polygons for which the perimeter is ≤ n.
 
-   For a given integer n, define P(n) as the number of distinct pythagorean
-   polygons for which the perimeter is ≤ n.
    Pythagorean polygons should be considered distinct as long as none is a
    translation of another.
 
@@ -7212,13 +7212,13 @@ Problem 294
    For a positive integer k, define d(k) as the sum of the digits of k in its
    usual decimal representation.Thus d(42) = 4+2 = 6.
 
-   For a positive integer n, define S(n) as the number of positive integers k
-   < 10^n with the following properties :
+For a positive integer n, define S(n) as the number of positive integers k
+< 10^n with the following properties :
 
-     • k is divisible by 23 and
-     • d(k) = 23.
+*   $\text k$ is divisible by 23 and
+*   $\text d(\text k) = 23$.
 
-   You are given that S(9) = 263626 and S(42) = 6377168878570056.
+You are given that S(9) = 263626 and S(42) = 6377168878570056.
 
    Find S(11^12) and give your answer mod 10^9.
 
@@ -7230,14 +7230,14 @@ Problem 295
 ===========
 
 
-   We call the convex area enclosed by two circles a lenticular hole if:
+We call the convex area enclosed by two circles a lenticular hole if:
 
-     • The centres of both circles are on lattice points.
-     • The two circles intersect at two distinct lattice points.
-     • The interior of the convex area enclosed by both circles does not
-       contain any lattice points.
+*   The centres of both circles are on lattice points.
+*   The two circles intersect at two distinct lattice points.
+*   The interior of the convex area enclosed by both circles does not contain any lattice points.
+
+Consider the circles:
 
-   Consider the circles:
    C[0]: x^2+y^2=25
    C[1]: (x+4)^2+(y-4)^2=1
    C[2]: (x-12)^2+(y-4)^2=65
@@ -7435,10 +7435,8 @@ If (n[1],n[2],n[3]) indicates a Nim position consisting of heaps of size
 n[1], n[2] and n[3] then there is a simple function X(n[1],n[2],n[3]) —
 that you may look up or attempt to deduce for yourself — that returns:
 
- • zero if, with perfect strategy, the player about to move will
-   eventually lose; or
- • non-zero if, with perfect strategy, the player about to move will
-   eventually win.
+*   zero if, with perfect strategy, the player about to move will eventually lose; or
+*   non-zero if, with perfect strategy, the player about to move will eventually win.
 
 For example X(1,2,3) = 0 because, no matter what the current player does,
 his opponent can respond with a move that leaves two heaps of equal size,
@@ -8029,8 +8027,8 @@ Let y[0], y[1], y[2],... be a sequence of random unsigned 32 bit integers
 
 For the sequence x[i] the following recursion is given:
 
-     • x[0] = 0 and
-     • x[i] = x[i-1] | y[i-1], for i > 0. ( | is the bitwise-OR operator)
+*   $x_0 = 0$ and
+*   $x_i = x_{i-1} | y_{i-1}$, for $i > 0$. ( `|` is the bitwise-OR operator)
 
 It can be seen that eventually there will be an index N such that x[i] =
 2^32 -1 (a bit-pattern of all ones) for all i ≥ N.
@@ -8473,8 +8471,8 @@ Problem 337
 
 Let {a[1], a[2],..., a[n]} be an integer sequence of length n such that:
 
-     • a[1] = 6
-     • for all 1 ≤ i < n : φ(a[i]) < φ(a[i+1]) < a[i] < a[i+1] ^1
+*   $a_1 = 6$
+*   for all $1 ≤ i < n$ : $φ(a_i) < φ(a_{i+1}) < a_i < {a_{i+1}}^1$
 
 Let S(N) be the number of such sequences with a[n] ≤ N.
 For example, S(10) = 4: {6}, {6, 8}, {6, 8, 9} and {6, 10}.
@@ -8847,51 +8845,49 @@ Problem 352
    Because of the high cost, the vet-in-charge suggests that instead of
    performing 25 separate tests, the following procedure can be used instead:
 
-   The sheep are split into 5 groups of 5 sheep in each group. For each
-   group, the 5 samples are mixed together and a single test is performed.
-   Then,
+The sheep are split into 5 groups of 5 sheep in each group. For each
+group, the 5 samples are mixed together and a single test is performed.
+Then,
+
+*   If the result is negative, all the sheep in that group are deemed to be virus-free.
+*   If the result is positive, 5 additional tests will be performed (a separate test for each animal) to determine the affected individual(s).
 
-     • If the result is negative, all the sheep in that group are deemed to
-       be virus-free.
-     • If the result is positive, 5 additional tests will be performed (a
-       separate test for each animal) to determine the affected
-       individual(s).
+Since the probability of infection for any specific animal is only 0.02,
+the first test (on the pooled samples) for each group will be:
 
-   Since the probability of infection for any specific animal is only 0.02,
-   the first test (on the pooled samples) for each group will be:
+*   Negative (and no more tests needed) with probability $0.98^5 = 0.9039207968$.
+*   Positive (5 additional tests needed) with probability $1 - 0.9039207968 = 0.0960792032$.
 
-     • Negative (and no more tests needed) with probability 0.98^5 =
-       0.9039207968.
-     • Positive (5 additional tests needed) with probability 1 - 0.9039207968
-       = 0.0960792032.
+Thus, the expected number of tests for each group is 1 + 0.0960792032 × 5
+= 1.480396016.
 
-   Thus, the expected number of tests for each group is 1 + 0.0960792032 × 5
-   = 1.480396016.
    Consequently, all 5 groups can be screened using an average of only
    1.480396016 × 5 = 7.40198008 tests, which represents a huge saving of more
    than 70% !
 
-   Although the scheme we have just described seems to be very efficient, it
-   can still be improved considerably (always assuming that the test is
-   sufficiently sensitive and that there are no adverse effects caused by
-   mixing different samples). E.g.:
-
-     • We may start by running a test on a mixture of all the 25 samples. It
-       can be verified that in about 60.35% of the cases this test will be
-       negative, thus no more tests will be needed. Further testing will only
-       be required for the remaining 39.65% of the cases.
-     • If we know that at least one animal in a group of 5 is infected and
-       the first 4 individual tests come out negative, there is no need to
-       run a test on the fifth animal (we know that it must be infected).
-     • We can try a different number of groups / different number of animals
-       in each group, adjusting those numbers at each level so that the total
-       expected number of tests will be minimised.
-
-   To simplify the very wide range of possibilities, there is one restriction
-   we place when devising the most cost-efficient testing scheme: whenever we
-   start with a mixed sample, all the sheep contributing to that sample must
-   be fully screened (i.e. a verdict of infected / virus-free must be reached
-   for all of them) before we start examining any other animals.
+Although the scheme we have just described seems to be very efficient, it
+can still be improved considerably (always assuming that the test is
+sufficiently sensitive and that there are no adverse effects caused by
+mixing different samples). E.g.:
+
+*   We may start by running a test on a mixture of all the 25 samples. It
+    can be verified that in about 60.35% of the cases this test will be
+    negative, thus no more tests will be needed. Further testing will only
+    be required for the remaining 39.65% of the cases.
+
+*   If we know that at least one animal in a group of 5 is infected and
+    the first 4 individual tests come out negative, there is no need to
+    run a test on the fifth animal (we know that it must be infected).
+
+*   We can try a different number of groups / different number of animals
+    in each group, adjusting those numbers at each level so that the total
+    expected number of tests will be minimised.
+
+To simplify the very wide range of possibilities, there is one restriction
+we place when devising the most cost-efficient testing scheme: whenever we
+start with a mixed sample, all the sheep contributing to that sample must
+be fully screened (i.e. a verdict of infected / virus-free must be reached
+for all of them) before we start examining any other animals.
 
    For the current example, it turns out that the most cost-efficient testing
    scheme (we'll call it the optimal strategy) requires an average of just
@@ -9511,18 +9507,16 @@ Problem 376
    die and have a larger than 50% chance of winning.
    A set of dice having this property is called a nontransitive set of dice.
 
-   We wish to investigate how many sets of nontransitive dice exist. We will
-   assume the following conditions:
+We wish to investigate how many sets of nontransitive dice exist. We will
+assume the following conditions:
 
-     • There are three six-sided dice with each side having between 1 and N
-       pips, inclusive.
-     • Dice with the same set of pips are equal, regardless of which side on
-       the die the pips are located.
-     • The same pip value may appear on multiple dice; if both players roll
-       the same value neither player wins.
-     • The sets of dice {A,B,C}, {B,C,A} and {C,A,B} are the same set.
+*   There are three six-sided dice with each side having between 1 and N pips, inclusive.
+*   Dice with the same set of pips are equal, regardless of which side on the die the pips are located.
+*   The same pip value may appear on multiple dice; if both players roll the same value neither player wins.
+*   The sets of dice {A,B,C}, {B,C,A} and {C,A,B} are the same set.
+
+For N = 7 we find there are 9780 such sets.
 
-   For N = 7 we find there are 9780 such sets.
    How many are there for N = 30 ?
 
    
@@ -9639,25 +9633,27 @@ Problem 382
    joined to form a closed chain or circuit. A polygon consists of at least
    three sides and does not self-intersect.
 
-   A set S of positive numbers is said to generate a polygon P if:
+A set S of positive numbers is said to generate a polygon P if:
 
-     • no two sides of P are the same length,
-     • the length of every side of P is in S, and
-     • S contains no other value.
+*   no two sides of P are the same length,
+*   the length of every side of P is in S, and
+*   S contains no other value.
+
+For example:
 
-   For example:
    The set {3, 4, 5} generates a polygon with sides 3, 4, and 5 (a triangle).
    The set {6, 9, 11, 24} generates a polygon with sides 6, 9, 11, and 24 (a
    quadrilateral).
    The sets {1, 2, 3} and {2, 3, 4, 9} do not generate any polygon at all.
 
-   Consider the sequence s, defined as follows:
+Consider the sequence s, defined as follows:
+
+*   $s_1 = 1, s_2 = 2, s_3 = 3$
+*   $s_n = s_{n-1} + s_{n-3}$ for $n > 3$.
 
-     • s[1] = 1, s[2] = 2, s[3] = 3
-     • s[n] = s[n-1] + s[n-3] for n > 3.
+Let U[n] be the set {s[1], s[2], ..., s[n]}. For example, U[10] = {1, 2,
+3, 4, 6, 9, 13, 19, 28, 41}.
 
-   Let U[n] be the set {s[1], s[2], ..., s[n]}. For example, U[10] = {1, 2,
-   3, 4, 6, 9, 13, 19, 28, 41}.
    Let f(n) be the number of subsets of U[n] which generate at least one
    polygon.
    For example, f(5) = 7, f(10) = 501 and f(25) = 18635853.
@@ -9913,23 +9909,18 @@ Problem 392
    gridlines does not have to be equidistant.
    An example of such grid is logarithmic graph paper.
 
-   Consider rectilinear grids in the Cartesian coordinate system with the
-   following properties:
+Consider rectilinear grids in the Cartesian coordinate system with the
+following properties:
 
-     • The gridlines are parallel to the axes of the Cartesian coordinate
-       system.
-     • There are N+2 vertical and N+2 horizontal gridlines. Hence there are
-       (N+1) x (N+1) rectangular cells.
-     • The equations of the two outer vertical gridlines are x = -1 and x =
-       1.
-     • The equations of the two outer horizontal gridlines are y = -1 and y =
-       1.
-     • The grid cells are colored red if they overlap with the unit circle,
-       black otherwise.
+*   The gridlines are parallel to the axes of the Cartesian coordinate system.
+*   There are $\text{N}+2$ vertical and $\text{N}+2$ horizontal gridlines. Hence there are $(\text{N}+1) \times (\text{N}+1)$ rectangular cells.
+*   The equations of the two outer vertical gridlines are $x = -1$ and $x = 1$.
+*   The equations of the two outer horizontal gridlines are $y = -1$ and $y = 1$.
+*   The grid cells are colored red if they overlap with the unit circle, black otherwise.
 
-   For this problem we would like you to find the postions of the remaining N
-   inner horizontal and N inner vertical gridlines so that the area occupied
-   by the red cells is minimized.
+For this problem we would like you to find the postions of the remaining N
+inner horizontal and N inner vertical gridlines so that the area occupied
+by the red cells is minimized.
 
    E.g. here is a picture of the solution for N = 10:
 
@@ -9999,14 +9990,16 @@ The Pythagorean tree is a fractal generated by the following procedure:
 Start with a unit square. Then, calling one of the sides its base (in the
 animation, the bottom side is the base):
 
-1. Attach a right triangle to the side opposite the base, with the
-   hypotenuse coinciding with that side and with the sides in a 3-4-5
-   ratio. Note that the smaller side of the triangle must be on the
-   'right' side with respect to the base (see animation).
-2. Attach a square to each leg of the right triangle, with one of its
-   sides coinciding with that leg.
-3. Repeat this procedure for both squares, considering as their bases the
-   sides touching the triangle.
+1.  Attach a right triangle to the side opposite the base, with the
+    hypotenuse coinciding with that side and with the sides in a 3-4-5
+    ratio. Note that the smaller side of the triangle must be on the
+    'right' side with respect to the base (see animation).
+
+2.  Attach a square to each leg of the right triangle, with one of its
+    sides coinciding with that leg.
+
+3.  Repeat this procedure for both squares, considering as their bases the
+    sides touching the triangle.
 
 The resulting figure, after an infinite number of iterations, is the
 Pythagorean tree.
@@ -10027,23 +10020,22 @@ Problem 396
 ===========
 
 
-   For any positive integer n, the nth weak Goodstein sequence {g[1], g[2],
-   g[3], ...} is defined as:
+For any positive integer n, the nth weak Goodstein sequence {g[1], g[2],
+g[3], ...} is defined as:
 
-     • g[1] = n
-     • for k > 1, g[k] is obtained by writing g[k-1] in base k, interpreting
-       it as a base k + 1 number, and subtracting 1.
+*   $g_1 = n$
+*   for $k > 1$, $g_k$ is obtained by writing $g_{k-1}$ in base $k$, interpreting it as a base $k + 1$ number, and subtracting 1.
 
-   The sequence terminates when g[k] becomes 0.
+The sequence terminates when g[k] becomes 0.
 
-   For example, the 6th weak Goodstein sequence is {6, 11, 17, 25, ...}:
+For example, the 6th weak Goodstein sequence is {6, 11, 17, 25, ...}:
 
-     • g[1] = 6.
-     • g[2] = 11 since 6 = 110[2], 110[3] = 12, and 12 - 1 = 11.
-     • g[3] = 17 since 11 = 102[3], 102[4] = 18, and 18 - 1 = 17.
-     • g[4] = 25 since 17 = 101[4], 101[5] = 26, and 26 - 1 = 25.
+*   $g_1 = 6$.
+*   $g_2 = 11$ since $6 = 110_2$, $110_3 = 12$, and $12 - 1 = 11$.
+*   $g_3 = 17$ since $11 = 102_3$, $102_4 = 18$, and $18 - 1 = 17$.
+*   $g_4 = 25$ since $17 = 101_4$, $101_5 = 26$, and $26 - 1 = 25$.
 
-   and so on.
+and so on.
 
    It can be shown that every weak Goodstein sequence terminates.
 
@@ -10136,15 +10128,16 @@ Problem 400
 ===========
 
 
-   A Fibonacci tree is a binary tree recursively defined as:
+A Fibonacci tree is a binary tree recursively defined as:
 
-     • T(0) is the empty tree.
-     • T(1) is the binary tree with only one node.
-     • T(k) consists of a root node that has T(k-1) and T(k-2) as children.
+*   $\text{T}(0)$ is the empty tree.
+*   $\text{T}(1)$ is the binary tree with only one node.
+*   $\text{T}(k)$ consists of a root node that has $\text{T}(k-1)$ and $\text{T}(k-2)$ as children.
+
+On such a tree two players play a take-away game. On each turn a player
+selects a node and removes that node along with the subtree rooted at that
+node.
 
-   On such a tree two players play a take-away game. On each turn a player
-   selects a node and removes that node along with the subtree rooted at that
-   node.
    The player who is forced to take the root node of the entire tree loses.
 
    Here are the winning moves of the first player on the first turn for T(k)
@@ -10278,18 +10271,16 @@ Problem 406
 ===========
 
 
-   We are trying to find a hidden number selected from the set of integers
-   {1, 2, ..., n} by asking questions. Each number (question) we ask, we get
-   one of three possible answers:
+We are trying to find a hidden number selected from the set of integers
+{1, 2, ..., n} by asking questions. Each number (question) we ask, we get
+one of three possible answers:
 
-     • "Your guess is lower than the hidden number" (and you incur a cost of
-       a), or
-     • "Your guess is higher than the hidden number" (and you incur a cost of
-       b), or
-     • "Yes, that's it!" (and the game ends).
+*   "Your guess is lower than the hidden number" (and you incur a cost of $a$), or
+*   "Your guess is higher than the hidden number" (and you incur a cost of $b$), or
+*   "Yes, that's it!" (and the game ends).
 
-   Given the value of n, a, and b, an optimal strategy minimizes the total
-   cost for the worst possible case.
+Given the value of n, a, and b, an optimal strategy minimizes the total
+cost for the worst possible case.
 
    For example, if n = 5, a = 2, and b = 3, then we may begin by asking "2"
    as our first question.
@@ -10372,16 +10363,16 @@ Problem 409
 ===========
 
 
-   Let n be a positive integer. Consider nim positions where:
+Let n be a positive integer. Consider nim positions where:
 
-     • There are n non-empty piles.
-     • Each pile has size less than 2^n.
-     • No two piles have the same size.
+*   There are n non-empty piles.
+*   Each pile has size less than $2^n$.
+*   No two piles have the same size.
 
-   Let W(n) be the number of winning nim positions satisfying the
-   aboveconditions (a position is winning if the first player has a winning
-   strategy). For example, W(1) = 1, W(2) = 6, W(3) = 168, W(5) = 19764360
-   and W(100) mod 1 000 000 007 = 384777056.
+Let W(n) be the number of winning nim positions satisfying the
+aboveconditions (a position is winning if the first player has a winning
+strategy). For example, W(1) = 1, W(2) = 6, W(3) = 168, W(5) = 19764360
+and W(100) mod 1 000 000 007 = 384777056.
 
    Find W(10 000 000) mod 1 000 000 007.
 
@@ -10511,16 +10502,15 @@ Problem 414
    E.g. base 15: (10,4,14,9,5)[15]
    base 21: (14,6,20,13,7)[21]
 
-   Define C[b] to be the Kaprekar constant in base b for 5 digits.Define the
-   function sb(i) to be
+Define C[b] to be the Kaprekar constant in base b for 5 digits.Define the
+function sb(i) to be
 
-     • 0 if i = C[b] or if i written in base b consists of 5 identical digits
-     • the number of iterations it takes the Kaprekar routine in base b to
-       arrive at C[b], otherwise
+*   0 if $i = C_b$ or if i written in base $b$ consists of 5 identical digits
+*   the number of iterations it takes the Kaprekar routine in base $b$ to arrive at $C_b$, otherwise
 
-   Note that we can define sb(i) for all integers i < b^5. If i written in
-   base b takes less than 5 digits, the number is padded with leading zero
-   digits until we have 5 digits before applying the Kaprekar routine.
+Note that we can define sb(i) for all integers i < b^5. If i written in
+base b takes less than 5 digits, the number is padded with leading zero
+digits until we have 5 digits before applying the Kaprekar routine.
 
    Define S(b) as the sum of sb(i) for 0 < i < b^5.
    E.g. S(15) = 5274369
@@ -10618,14 +10608,14 @@ Problem 418
 ===========
 
 
-   Let n be a positive integer. An integer triple (a, b, c) is called a
-   factorisation triple of n if:
+Let n be a positive integer. An integer triple (a, b, c) is called a
+factorisation triple of n if:
 
-     • 1 ≤ a ≤ b ≤ c
-     • a·b·c = n.
+*   $1 ≤ a ≤ b ≤ c$
+*   $a·b·c = n$.
 
-   Define f(n) to be a + b + c for the factorisation triple (a, b, c) of n
-   which minimises c / a. One can show that this triple is unique.
+Define f(n) to be a + b + c for the factorisation triple (a, b, c) of n
+which minimises c / a. One can show that this triple is unique.
 
    For example, f(165) = 19, f(100100) = 142 and f(20!) = 4034872.
 
@@ -10712,17 +10702,14 @@ Problem 422
 
    Next, define X as the point (7, 1). It can be seen that X is in H.
 
-   Now we define a sequence of points in H, {P[i] : i ≥ 1}, as:
+Now we define a sequence of points in H, {P[i] : i ≥ 1}, as:
 
-     • P[1] = (13, 61/4).
-     • P[2] = (-43/6, -4).
-     • For i > 2, P[i] is the unique point in H that is different from P[i-1]
-       and such that line P[i]P[i-1] is parallel to line P[i-2]X. It can be
-       shown that P[i] is well-defined, and that its coordinates are always
-       rational.
+*   $\text{P}_1 = (13, 61/4)$.
+*   $\text{P}_2 = (-43/6, -4)$.
+*   For $i > 2$, $\text{P}_i$ is the unique point in $\text{H}$ that is different from $\text{P}_{i-1}$ and such that line $\text{P}_i\text{P}_{i-1}$ is parallel to line $\text{P}_{i-2}\text{X}$. It can be shown that $\text{P}_i$ is well-defined, and that its coordinates are always rational.
 
-   You are given that P[3] = (-19/2, -229/24), P[4] = (1267/144, -37/12) and
-   P[7] = (17194218091/143327232, 274748766781/1719926784).
+You are given that P[3] = (-19/2, -229/24), P[4] = (1267/144, -37/12) and
+P[7] = (17194218091/143327232, 274748766781/1719926784).
 
    Find P[n] for n = 11^14 in the following format:
    If P[n] = (a/b, c/d) where the fractions are in lowest terms and the
@@ -10880,9 +10867,9 @@ evolves to [1, 2, 3]; we shall call this the final state.
 
 We define the sequence {t[i]}:
 
- • s[0] = 290797
- • s[k+1] = s[k]^2 mod 50515093
- • t[k] = (s[k] mod 64) + 1
+*   $s_0 = 290797$
+*   $s_{k+1} = {s_k}^2 \bmod 50515093$
+*   $t_k = (s_k \bmod 64) + 1$
 
 Starting from the initial configuration (t[0], t[1], …, t[10]), the final
 state becomes [1, 3, 10, 24, 51, 75].
@@ -10928,17 +10915,16 @@ Problem 428
    Let C[in] be the circle having the diameter XY.
    Let C[out] be the circle having the diameter WZ.
 
-   The triplet (a, b, c) is called a necklace triplet if you can place k ≥ 3
-   distinct circles C[1], C[2], ..., C[k] such that:
+The triplet (a, b, c) is called a necklace triplet if you can place k ≥ 3
+distinct circles C[1], C[2], ..., C[k] such that:
 
-     • C[i] has no common interior points with any C[j] for 1 ≤ i, j ≤ k and
-       i ≠ j,
-     • C[i] is tangent to both C[in] and C[out] for 1 ≤ i ≤ k,
-     • C[i] is tangent to C[i+1] for 1 ≤ i < k, and
-     • C[k] is tangent to C[1].
+*   $C_i$ has no common interior points with any $C_j$ for $1 ≤ i$, $j ≤ k$ and $i ≠ j$,
+*   $C_i$ is tangent to both $C_{in}$ and $C_{out}$ for $1 ≤ i ≤ k$,
+*   $C_i$ is tangent to $C_{i+1}$ for $1 ≤ i < k$, and
+*   $C_k$ is tangent to $C_1$.
 
-   For example, (5, 5, 5) and (4, 3, 21) are necklace triplets, while it can
-   be shown that (2, 2, 5) is not.
+For example, (5, 5, 5) and (4, 3, 21) are necklace triplets, while it can
+be shown that (2, 2, 5) is not.
 
    Let T(n) be the number of necklace triplets (a, b, c) such that a, b and c
    are positive integers, and b ≤ n.For example, T(1) = 9, T(20) = 732 and
@@ -11210,14 +11196,14 @@ Problem 438
    the solutions of the polynomial equation x^n + a[1]x^n-1 + a[2]x^n-2 + ...
    + a[n-1]x + a[n] = 0.
 
-   Consider the following two conditions:
+Consider the following two conditions:
 
-     • x[1], ..., x[n] are all real.
-     • If x[1], ..., x[n] are sorted, ⌊x[i]⌋ = i for 1 ≤ i ≤ n. (⌊·⌋: floor
-       function.)
+*   $x_1, ..., x_n$ are all real.
+*   If $x_1, ..., x_n$ are sorted, $⌊x_i⌋ = i$ for $1 ≤ i ≤ n$. (⌊·⌋: floor function.)
+
+In the case of n = 4, there are 12 n-tuples of integers which satisfy both
+conditions.
 
-   In the case of n = 4, there are 12 n-tuples of integers which satisfy both
-   conditions.
    We define S(t) as the sum of the absolute values of the integers in t.
    For n = 4 we can verify that ∑S(t) = 2087 for all n-tuples t which satisfy
    both conditions.
@@ -11276,14 +11262,15 @@ Problem 441
 ===========
 
 
-   For an integer M, we define R(M) as the sum of 1/(p·q) for all the integer
-   pairs p and q which satisfy all of these conditions:
+For an integer M, we define R(M) as the sum of 1/(p·q) for all the integer
+pairs p and q which satisfy all of these conditions:
+
+*   $1 ≤ p < q ≤ M$
+*   $p + q ≥ M$
+*   $p$ and $q$ are coprime.
 
-     • 1 ≤ p < q ≤ M
-     • p + q ≥ M
-     • p and q are coprime.
+We also define S(N) as the sum of R(i) for 2 ≤ i ≤ N.
 
-   We also define S(N) as the sum of R(i) for 2 ≤ i ≤ N.
    We can verify that S(2) = R(2) = 1/2, S(10) ≈ 6.9147 and S(100) ≈ 58.2962.
 
    Find S(10^7). Give your answer rounded to four decimal places.
@@ -11617,14 +11604,14 @@ Problem 455
    last 9 digits of n^x form the number x (including leading zeros), or zero
    if no such integer exists.
 
-   For example:
+For example:
 
-     • f(4) = 411728896 (4^411728896 = ...490411728896)
-     • f(10) = 0
-     • f(157) = 743757 (157^743757 = ...567000743757)
-     • Σf(n), 2 ≤ n ≤ 10^3 = 442530011399
+*   $f(4) = 411728896 (4^{411728896} = ...490411728896)$
+*   $f(10) = 0$
+*   $f(157) = 743757 (157^{743757} = ...567000743757)$
+*   $Σf(n), 2 ≤ n ≤ 10^3 = 442530011399$
 
-   Find Σf(n), 2 ≤ n ≤ 10^6.
+Find Σf(n), 2 ≤ n ≤ 10^6.
 
    
    Answer: 22d6cf30a29e14e5c78dca980edc2796
@@ -11700,14 +11687,14 @@ Problem 459
    Each square contains a disk with one side white and one side black.
    The game starts with all disks showing their white side.
 
-   A turn consists of flipping all disks in a rectangle with the following
-   properties:
+A turn consists of flipping all disks in a rectangle with the following
+properties:
 
-     • the upper right corner of the rectangle contains a white disk
-     • the rectangle width is a perfect square (1, 4, 9, 16, ...)
-     • the rectangle height is a triangular number (1, 3, 6, 10, ...)
+*   the upper right corner of the rectangle contains a white disk
+*   the rectangle width is a perfect square (1, 4, 9, 16, ...)
+*   the rectangle height is a triangular number (1, 3, 6, 10, ...)
 
-   Players alternate turns. A player wins by turning the grid all black.
+Players alternate turns. A player wins by turning the grid all black.
 
    Let W(N) be the number of winning moves for the first player on a N by N
    board with all disks white, assuming perfect play.
@@ -11725,21 +11712,21 @@ Problem 460
 ===========
 
 
-   On the Euclidean plane, an ant travels from point A(0, 1) to point B(d, 1)
-   for an integer d.
+On the Euclidean plane, an ant travels from point A(0, 1) to point B(d, 1)
+for an integer d.
 
-   In each step, the ant at point (x[0], y[0]) chooses one of the lattice
-   points (x[1], y[1]) which satisfy x[1] ≥ 0 and y[1] ≥ 1 and goes straight
-   to (x[1], y[1]) at a constant velocity v. The value of v depends on y[0]
-   and y[1] as follows:
+In each step, the ant at point (x[0], y[0]) chooses one of the lattice
+points (x[1], y[1]) which satisfy x[1] ≥ 0 and y[1] ≥ 1 and goes straight
+to (x[1], y[1]) at a constant velocity v. The value of v depends on y[0]
+and y[1] as follows:
 
-     • If y[0] = y[1], the value of v equals y[0].
-     • If y[0] ≠ y[1], the value of v equals (y[1] - y[0]) / (ln(y[1]) -
-       ln(y[0])).
+*   If $y_0 = y_1$, the value of $v$ equals $y_0$.
+*   If $y_0 ≠ y_1$, the value of $v$ equals $(y_1 - y_0) / (\ln(y_1) - \ln(y_0))$.
+
+The left image is one of the possible paths for d = 4. First the ant goes
+from A(0, 1) to P[1](1, 3) at velocity (3 - 1) / (ln(3) - ln(1)) ≈ 1.8205.
+Then the required time is sqrt(5) / 1.8205 ≈ 1.2283.
 
-   The left image is one of the possible paths for d = 4. First the ant goes
-   from A(0, 1) to P[1](1, 3) at velocity (3 - 1) / (ln(3) - ln(1)) ≈ 1.8205.
-   Then the required time is sqrt(5) / 1.8205 ≈ 1.2283.
    From P[1](1, 3) to P[2](3, 3) the ant travels at velocity 3 so the
    required time is 2 / 3 ≈ 0.6667. From P[2](3, 3) to B(4, 1) the ant
    travels at velocity (1 - 3) / (ln(1) - ln(3)) ≈ 1.8205 so the required
@@ -11816,15 +11803,15 @@ Problem 463
 ===========
 
 
-   The function $f$ is defined for all positive integers as follows:
+The function $f$ is defined for all positive integers as follows:
 
-     • $f(1)=1$
-     • $f(3)=3$
-     • $f(2n)=f(n)$
-     • $f(4n + 1)=2f(2n + 1) - f(n)$
-     • $f(4n + 3)=3f(2n + 1) - 2f(n)$
+*   $f(1) = 1$
+*   $f(3) = 3$
+*   $f(2n) = f(n)$
+*   $f(4n + 1) = 2f(2n + 1) - f(n)$
+*   $f(4n + 3) = 3f(2n + 1) - 2f(n)$
 
-   The function $S(n)$ is defined as $\sum_{i=1}^{n}f(i)$.
+The function $S(n)$ is defined as $\sum_{i=1}^{n}f(i)$.
 
    $S(8)=22$ and $S(100)=3604$.
 
@@ -11838,25 +11825,25 @@ Problem 464
 ===========
 
 
-   The Möbius function, denoted μ(n), is defined as:
+The Möbius function, denoted μ(n), is defined as:
+
+*   $μ(n) = (-1)^{ω(n)}$ if $n$ is squarefree (where $ω(n)$ is the number of distinct prime factors of $n$)
+*   $μ(n) = 0$ if $n$ is not squarefree.
 
-     • μ(n) = (-1)^ω(n) if n is squarefree (where ω(n) is the number of
-       distinct prime factors of n)
-     • μ(n) = 0 if n is not squarefree.
+Let P(a,b) be the number of integers n in the interval [a,b] such that
+μ(n) = 1.
 
-   Let P(a,b) be the number of integers n in the interval [a,b] such that
-   μ(n) = 1.
    Let N(a,b) be the number of integers n in the interval [a,b] such that
    μ(n) = -1.
    For example, P(2,10) = 2 and N(2,10) = 4.
 
-   Let C(n) be the number of integer pairs (a,b) such that:
+Let C(n) be the number of integer pairs (a,b) such that:
 
-     • 1 ≤ a ≤ b ≤ n,
-     • 99·N(a,b) ≤ 100·P(a,b), and
-     • 99·P(a,b) ≤ 100·N(a,b).
+*   $1 ≤ a ≤ b ≤ n$,
+*   $99·\text{N}(a,b) ≤ 100·\text{P}(a,b)$, and
+*   $99·\text{P}(a,b) ≤ 100·\text{N}(a,b)$.
 
-   For example, C(10) = 13, C(500) = 16676 and C(10 000) = 20155319.
+For example, C(10) = 13, C(500) = 16676 and C(10 000) = 20155319.
 
    Find C(20 000 000).
 
@@ -12087,19 +12074,21 @@ Problem 472
 ===========
 
 
-   There are N seats in a row. N people come one after another to fill the
-   seats according to the following rules:
+There are N seats in a row. N people come one after another to fill the
+seats according to the following rules:
+
+1.  No person sits beside another.
+
+2.  The first person chooses any seat.
 
-    1. No person sits beside another.
-    2. The first person chooses any seat.
-    3. Each subsequent person chooses the seat furthest from anyone else
-       already seated, as long as it does not violate rule 1. If there is
-       more than one choice satisfying this condition, then the person
-       chooses the leftmost choice.
+3.  Each subsequent person chooses the seat furthest from anyone else
+    already seated, as long as it does not violate rule 1. If there is
+    more than one choice satisfying this condition, then the person
+    chooses the leftmost choice.
 
-   Note that due to rule 1, some seats will surely be left unoccupied, and
-   the maximum number of people that can be seated is less than N (for N >
-   1).
+Note that due to rule 1, some seats will surely be left unoccupied, and
+the maximum number of people that can be seated is less than N (for N >
+1).
 
    Here are the possible seating arrangements for N = 15: