Also known as Project Euler problem 39:

If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.

{20,48,52}, {24,45,51}, {30,40,50}

For which value of p ≤ 1000, is the number of solutions maximised?

My first thought here was to generate an iterable collection of right angled triangles and accumulate a count of triangles by perimeter size. Don’t try this – it is far too slow an approach.

So my second thought was to try and iterate over the perimeter sizes and calculate the number of right angle triangles for each. It’s a long time since I had to think about Pythagoras but, for a triangle with sides a, b and c and a perimeter of p, I can make the following two statements:

a + b + c = p
a^2 + b^2 = c^2

Based on the second statement, above, I can also assert:
c = sqrt(a^2 + b^2)

Also:
c = p - a - b

Which means:
a^2 + b^2 = (p - a - b)^2

This can be expanded and simplified to state:
0 = pp - 2pb - 2ap + 2ab

or:

b = p(p - 2a) / 2(p - a)

Using this formula, we can say that for any values of p and a where p > a, if b is an integer then we have a right angle triangle.

One further optimisation: Intuitively, a can’t be greater than p/3.

Implementing this can then be done with two very simple for loops. The outer one iterates over all values of p and the inner one iterates over the range (2, int(p / 3)) and counts the number of integer values of b.

Also known as Project Euler Problem 38:

Take the number 192 and multiply it by each of 1, 2, and 3:

192 × 1 = 192
192 × 2 = 384
192 × 3 = 576

By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)

The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, … , n) where n > 1?

This is a problem for which finding a practical approach took a lot longer than the implementation of said approach. It’s also a problem that forced me to think about the actual maths of the problem. I already have an is_pandigital function, which I wrote to help solve problem 32, so the main challenge here is to constrain the search enough that I can avoid endless looping.

So, a bit of analysis…

For n = 2,
9 * (1, 2) = 918
98 * (1, 2) = 98196
987 * (1, 2) = 9871974
9876 * (1, 2) = 987619752 <- That's 9 digits

Since the problem mentions the pandigital, 918273645, I plugged this into the calculator as well,
9182 * (1, 2) = 918218365 <- That's 9 digits again

For n = 3,
9 * (1, 2, 3) = 91827
98 * (1, 2, 3) = 98196294 <- 8 digits is not enough
987 * (1, 2, 3) = 98719742961 <- 11 digits is too many

So I now know that the number I am looking for is n * (1, 2) where n is between 9876 and 9182.

The implementation, therefore, is a very simple loop that checks is_pandigital(int(str(n) + str(n * 2))) for values of n from 9876 to 9182.

The result is almost instantaneous.