Syntax fixes 0-100

project_euler/problems/004_largest_palindrome_product.txt

@@ -2,6 +2,6 @@ http://projecteuler.net/problem=004
 
 Largest palindrome product
 
-A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91  99.
+A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 x 99.
 Find the largest palindrome made from the product of two 3-digit numbers.
 

project_euler/problems/006_sum_square_difference.txt

@@ -6,6 +6,6 @@ The sum of the squares of the first ten natural numbers is,
 12 + 22 + ... + 102 = 385
 The square of the sum of the first ten natural numbers is,
 (1 + 2 + ... + 10)2 = 552 = 3025
-Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025  385 = 2640.
+Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 - 385 = 2640.
 Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
 

project_euler/problems/009_special_pythagorean_triplet.txt

@@ -2,7 +2,7 @@ http://projecteuler.net/problem=009
 
 Special Pythagorean triplet
 
-A Pythagorean triplet is a set of three natural numbers, a  b  c, for which,
+A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
  a2 + b2 = c2
 For example, 32 + 42 = 9 + 16 = 25 = 52.
 There exists exactly one Pythagorean triplet for which a + b + c = 1000.Find the product abc.

project_euler/problems/020_factorial_digit_sum.txt

@@ -2,7 +2,7 @@ http://projecteuler.net/problem=020
 
 Factorial digit sum
 
-n! means n  (n  1)  ...  3  2  1
-For example, 10! = 10  9  ...  3  2  1 = 3628800,and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
+n! means n x (n − 1) x ... x 3 x 2 x 1
+For example, 10! = 10 x 9 x ... x 3 x 2 x 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
 Find the sum of the digits in the number 100!
 

project_euler/problems/021_amicable_numbers.txt

@@ -3,7 +3,7 @@ http://projecteuler.net/problem=021
 Amicable numbers
 
 Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
-If d(a) = b and d(b) = a, where a  b, then a and b are an amicable pair and each of a and b are called amicable numbers.
+If d(a) = b and d(b) = a, where a <> b, then a and b are an amicable pair and each of a and b are called amicable numbers.
 For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
 Evaluate the sum of all the amicable numbers under 10000.
 

project_euler/problems/022_names_scores.txt

@@ -3,6 +3,6 @@ http://projecteuler.net/problem=022
 Names scores
 
 Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.
-For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938  53 = 49714.
+For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 x 53 = 49714.
 What is the total of all the name scores in the file?
 

project_euler/problems/025_1000_digit_fibonacci_number.txt

@@ -5,17 +5,17 @@ http://projecteuler.net/problem=025
 The Fibonacci sequence is defined by the recurrence relation:
 Fn = Fn1 + Fn2, where F1 = 1 and F2 = 1.
 Hence the first 12 terms will be:
-F1 = 1
-F2 = 1
-F3 = 2
-F4 = 3
-F5 = 5
-F6 = 8
-F7 = 13
-F8 = 21
-F9 = 34
-F10 = 55
-F11 = 89
+F1 = 1
+F2 = 1
+F3 = 2
+F4 = 3
+F5 = 5
+F6 = 8
+F7 = 13
+F8 = 21
+F9 = 34
+F10 = 55
+F11 = 89
 F12 = 144
 The 12th term, F12, is the first term to contain three digits.
 What is the first term in the Fibonacci sequence to contain 1000 digits?

project_euler/problems/046_goldbachs_other_conjecture.txt

@@ -3,12 +3,12 @@ http://projecteuler.net/problem=046
 Goldbach's other conjecture
 
 It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.
-9 = 7 + 212
-15 = 7 + 222
-21 = 3 + 232
-25 = 7 + 232
-27 = 19 + 222
-33 = 31 + 212
+9 = 7 + 2 x 12
+15 = 7 + 2 x 22
+21 = 3 + 2 x 32
+25 = 7 + 2 x 32
+27 = 19 + 2 x 22
+33 = 31 + 2 x 12
 It turns out that the conjecture was false.
 What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
 

project_euler/problems/057_square_root_convergents.txt

@@ -3,11 +3,11 @@ http://projecteuler.net/problem=057
 Square root convergents
 
 It is possible to show that the square root of two can be expressed as an infinite continued fraction.
- 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
+√ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
 By expanding this for the first four iterations, we get:
-1 + 1/2 = 3/2 = 1.5
-1 + 1/(2 + 1/2) = 7/5 = 1.4
-1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
+1 + 1/2 = 3/2 = 1.5
+1 + 1/(2 + 1/2) = 7/5 = 1.4
+1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
 1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...
 The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
 In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?

project_euler/problems/061_cyclical_figurate_numbers.txt

@@ -21,28 +21,28 @@ P4,n=n2
 
 Pentagonal
 
-P5,n=n(3n1)/2
+P5,n=n(3n-1)/2
 
 1, 5, 12, 22, 35, ...
 
 
 Hexagonal
 
-P6,n=n(2n1)
+P6,n=n(2n-1)
 
 1, 6, 15, 28, 45, ...
 
 
 Heptagonal
 
-P7,n=n(5n3)/2
+P7,n=n(5n-3)/2
 
 1, 7, 18, 34, 55, ...
 
 
 Octagonal
 
-P8,n=n(3n2)
+P8,n=n(3n-2)
 
 1, 8, 21, 40, 65, ...
 

project_euler/problems/064_odd_period_square_roots.txt

@@ -6,7 +6,7 @@ All square roots are periodic when written as continued fractions and can be wri
 
 
 
-N = a0 +
+√N = a0 +
 1
 
 
@@ -28,29 +28,29 @@ a3 + ...
 
 
 
-For example, let us consider 23:
+For example, let us consider √23:
 
 
 
-23 = 4 + 23 — 4 = 4 + 
+√23 = 4 + √23 — 4 = 4 + 
 1
  = 4 + 
 1
 
 
 
-123—4
+1√23—4
 
 1 + 
-23 – 37
+√23 – 37
 
 
 
 If we continue we would get the following expansion:
 
 
 
-23 = 4 +
+√23 = 4 +
 1
 
 
@@ -86,89 +86,89 @@ The process can be summarised as follows:
 
 a0 = 4,
 
-123—4
+1√23—4
  = 
-23+47
+√23+47
  = 1 + 
-23—37
+√23—37
 
 
 a1 = 1,
 
-723—3
+7√23—3
  = 
-7(23+3)14
+7(√23+3)14
  = 3 + 
-23—32
+√23—32
 
 
 a2 = 3,
 
-223—3
+2√23—3
  = 
-2(23+3)14
+2(√23+3)14
  = 1 + 
-23—47
+√23—47
 
 
 a3 = 1,
 
-723—4
+7√23—4
  = 
-7(23+4)7
+7(√23+4)7
  = 8 + 
-23—4
+√23—4
 
 
 a4 = 8,
 
-123—4
+1√23—4
  = 
-23+47
+√23+47
  = 1 + 
-23—37
+√23—37
 
 
 a5 = 1,
 
-723—3
+7√23—3
  = 
-7(23+3)14
+7(√23+3)14
  = 3 + 
-23—32
+√23—32
 
 
 a6 = 3,
 
-223—3
+2√23—3
  = 
-2(23+3)14
+2(√23+3)14
  = 1 + 
-23—47
+√23—47
 
 
 a7 = 1,
 
-723—4
+7√23—4
  = 
-7(23+4)7
+7(√23+4)7
  = 8 + 
-23—4
+√23—4
 
 
 
-It can be seen that the sequence is repeating. For conciseness, we use the notation 23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.
+It can be seen that the sequence is repeating. For conciseness, we use the notation √23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.
 The first ten continued fraction representations of (irrational) square roots are:
-2=[1;(2)], period=1
-3=[1;(1,2)], period=2
-5=[2;(4)], period=1
-6=[2;(2,4)], period=2
-7=[2;(1,1,1,4)], period=4
-8=[2;(1,4)], period=2
-10=[3;(6)], period=1
-11=[3;(3,6)], period=2
-12= [3;(2,6)], period=2
-13=[3;(1,1,1,1,6)], period=5
-Exactly four continued fractions, for N  13, have an odd period.
-How many continued fractions for N  10000 have an odd period?
+√2=[1;(2)], period=1
+√3=[1;(1,2)], period=2
+√5=[2;(4)], period=1
+√6=[2;(2,4)], period=2
+√7=[2;(1,1,1,4)], period=4
+√8=[2;(1,4)], period=2
+√10=[3;(6)], period=1
+√11=[3;(3,6)], period=2
+√12= [3;(2,6)], period=2
+√13=[3;(1,1,1,1,6)], period=5
+Exactly four continued fractions, for N <= 13, have an odd period.
+How many continued fractions for N <= 10000 have an odd period?
 

project_euler/problems/065_convergents_of_e.txt

@@ -6,7 +6,7 @@ The square root of 2 can be written as an infinite continued fraction.
 
 
 
-2 = 1 +
+√2 = 1 +
 1
 
 
@@ -36,8 +36,8 @@ The square root of 2 can be written as an infinite continued fraction.
 
 
 
-The infinite continued fraction can be written, 2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, 23 = [4;(1,3,1,8)].
-It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for 2.
+The infinite continued fraction can be written, √2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, √23 = [4;(1,3,1,8)].
+It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for √2.
 
 
 
@@ -132,7 +132,7 @@ It turns out that the sequence of partial values of continued fractions for squa
 
 
 
-Hence the sequence of the first ten convergents for 2 are:
+Hence the sequence of the first ten convergents for √2 are:
 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...
 What is most surprising is that the important mathematical constant,e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].
 The first ten terms in the sequence of convergents for e are:

project_euler/problems/066_diophantine_equation.txt

@@ -3,15 +3,14 @@ http://projecteuler.net/problem=066
 Diophantine equation
 
 Consider quadratic Diophantine equations of the form:
-x2 – Dy2 = 1
-For example, when D=13, the minimal solution in x is 6492 – 131802 = 1.
+x2 - Dy2 = 1
+For example, when D=13, the minimal solution in x is 6492 - 13 x 1802 = 1.
 It can be assumed that there are no solutions in positive integers when D is square.
 By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:
-32 – 222 = 1
-22 – 312 = 1
-92 – 542 = 1
-52 – 622 = 1
-82 – 732 = 1
-Hence, by considering minimal solutions in x for D  7, the largest x is obtained when D=5.
-Find the value of D  1000 in minimal solutions of x for which the largest value of x is obtained.
+32 - 2 x 22 = 1
+22 - 3 x 12 = 192 - 5 x 42 = 1
+52 - 6 x 22 = 1
+82 - 7 x 32 = 1
+Hence, by considering minimal solutions in x for D <= 7, the largest x is obtained when D=5.
+Find the value of D <= 1000 in minimal solutions of x for which the largest value of x is obtained.
 

project_euler/problems/069_totient_maximum.txt

@@ -67,6 +67,6 @@ n/φ(n)
 
 
 
-It can be seen that n=6 produces a maximum n/φ(n) for n  10.
-Find the value of n  1,000,000 for which n/φ(n) is a maximum.
+It can be seen that n=6 produces a maximum n/φ(n) for n <= 10.
+Find the value of n <= 1,000,000 for which n/φ(n) is a maximum.
 

project_euler/problems/070_totient_permutation.txt

@@ -4,5 +4,5 @@ Totient permutation
 
 Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of positive numbers less than or equal to n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.The number 1 is considered to be relatively prime to every positive number, so φ(1)=1. 
 Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180.
-Find the value of n, 1  n  107, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.
+Find the value of n, 1 < n < 107, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.
 

project_euler/problems/071_ordered_fractions.txt

@@ -2,7 +2,7 @@ http://projecteuler.net/problem=071
 
 Ordered fractions
 
-Consider the fraction, n/d, where n and d are positive integers. If nd and HCF(n,d)=1, it is called a reduced proper fraction.
+Consider the fraction, n/d, where n and d are positive integers. If n<=d and HCF(n,d)=1, it is called a reduced proper fraction.
 If we list the set of reduced proper fractions for d <= 8 in ascending order of size, we get:
 
 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8