$\{20,48,52\}, \{24,45,51\}, \{30,40,50\}$
For which value of p ≤ 1000, is the number of solutions maximised?
+ + + +We are looking for right angle triangles so Pythagoras' theorem $a^2 + b^2 = c^2$ can be used. Also it must be true that $a <= b <= c$ is true for every solution. Let's start with a function that tests the given examples.
+
+def is_right_angle_triangle(a, b, c, p):
+ if a + b + c != p:
+ return False
+ if a**2 + b**2 != c**2:
+ return False
+ return True
+
+given = [(20, 48, 52, 120), (24, 45, 51, 120), (30, 40, 50, 120)]
+for g in given:
+ assert(is_right_angle_triangle(*g))
+assert(is_right_angle_triangle(29, 41, 50, 120) == False)
This seems bruteforceable. Let's try to find the solutions for p = 120.
+ +def brute_force(p):
+ solutions = [(a,b, p - a - b)
+ for b in range(1, p // 2 + 1)
+ for a in range(1, b)
+ if a*a + b*b == (p - a - b) * (p - a - b)
+ ]
+ return solutions
+
+print(brute_force(120))
+Looks good. Let's try for all $p <= 1000$.
+ +solutions = [(len(brute_force(p)), p) for p in range(1, 1001)]
+Well, not too fast, but also not too slow. Let's get the max and we have our solution.
+ +s, p = max(solutions)
+print(p)
+assert(p == 840)
+We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
+What is the largest n-digit pandigital prime that exists?
+ +
+The nth term of the sequence of triangle numbers is given by, $t_n = \frac{1}{2}n(n+1)$; so the first ten triangle numbers are:
+ +1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
+
+By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = $t_{10}$. If the word value is a triangle number then we shall call the word a triangle word.
+Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, how many are triangle words? (The file is saved in the same directory as this notebook file as EulerProblem042.txt)
+ +
+The number, 1406357289, is a 0 to 9 pandigital number because it is made 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:
+$d_2d_3d_4$=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.
+ +
+