# Project Euler – Problem 64 Solution

#### Problem

All square roots are periodic when written as continued fractions and can be written in the form:

For example, let us consider ?23:

If we continue we would get the following expansion:

The process can be summarised as follows:

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?

#### Solution

(see full solution here).

Based on the algorithm on continued fractions from Wikipedia, we can implement the expansion algorithm as:

From the Wikipedia page:

The algorithm can also terminate on ai when ai = 2 a0, which is easier to implement.

which corresponds to the termination condition we have in the Repeat active pattern (which also checks if the accumulator is empty):

Also, this algorithm doesn’t work on numbers that are perfect squares, i.e. 4, 9, 16, … hence we need to exclude them when searching for our answer.

Here’s the solution in full:

This solution took 92ms to execute on my machine.

# Project Euler – Problem 80 Solution

#### Problem

It is well known that if the square root of a natural number is not an integer, then it is irrational. The decimal expansion of such square roots is infinite without any repeating pattern at all.

The square root of two is 1.41421356237309504880…, and the digital sum of the first one hundred decimal digits is 475.

For the first one hundred natural numbers, find the total of the digital sums of the first one hundred decimal digits for all the irrational square roots.

#### Solution

(see full solution here).

The premise of the problem itself is fairly straight forward, the challenge here is down to the way floating points are implemented on computers which lacks the precision necessary to solve this problem. So the bulk of the research went into finding a way to generate an arbitrary number of digits of a square root.

As usual, Wikipedia has plenty to offer and the easiest solution implementation wise is this solution by Frazer Jarvis.

which translates to:

The rest is really straight forward, with the only tricky thing being the conversion from char to int since this returns the internal integer value instead – e.g. int ‘0’ => 48 and int ‘1’ => 49 – hence the need for some hackery in the sum function.

Here is the full solution:

The solution took 95ms to complete on my machine.

# Project Euler – Problem 61 Solution

#### Problem

Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:

The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.

1. The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).
2. Each polygonal type: triangle (P3,127=8128), square (P4,91=8281), and pentagonal (P5,44=2882), is represented by a different number in the set.
3. This is the only set of 4-digit numbers with this property.

Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.

#### Solution

(see full solution here),

The tricky thing here (at least for me) was to remember that the six 4-digit numbers have to come from different sets, but not necessarily in the order of P3, P4, … P8. Once that is cleared up, the rest is fairly straight forward. In the solution linked above, I first created a set of functions for generating triangle, square, pentagonal, … octagonal numbers:

Since the question concerns only 4-digit numbers, so for efficiency sake let’s generate the desired 4 digit numbers ahead of time and safe them for later use:

The is4digit predicate function is self-explanatory. naturalNumbers is an infinite sequence of integers starting from 1, we use this sequence to generate the figurate numbers we need, but only keep those that are actually 4 digits.

So far so good, we have all the figurate numbers in an array where [0] => P3, [1] => P4, and so on.

Next, create permutations of the figurate numbers such that we exhaust all possible sequence of figurate numbers:

P3 => P4 => P5 => P6 => P7 => P8

P3 => P4 => P6 => P7 => P8 => P5

P4 => P3 => P5 => P6 => P7 => P8

(P.S. the permute function here comes from the Common.fs source file in the solution)

To find the answer to the problem, we process each permutation to find our answer, take a moment to understand this code:

The processPermutation function processes one permutation of the figurate numbers and the termination conditions for the inner loop function are:

1. we have one number from each figurate number set and that the last 2 digits of the last number = first 2 digits of first number

2. we have one number from each figurate number set but last 2 digits of last number <> first 2 digits of first number (so close!)

3. one of the figurate number set in the sequence doesn’t contain a number whose first 2 digits = the last 2 digits of the number selected from the last figurate number set (short-circuited)

For each number in the current set of figurate numbers we build up a new predicate function – e.g. if x = 1282 then the predicate function would find 4-digit numbers whose first two digit = 82 – and use it to process the next set of figurate numbers in the sequence.

The loop function returns int list option where the int list represents the cyclic figurate numbers we’re looking for, so all that’s left is to unpack the option type and then sum the list.

This solution took 17ms to find the solution on my machine.

# Project Euler – Problem 65 Solution

#### Problem

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.

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:

2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, …

The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17.

Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.

#### Solution

If you look at the convergents of ?2 and the numerators in the convergents of e, you’ll see a pattern emerging:

1    +    2    *   1    =    3

2    +    3    *   2    =    8

3    +    8    *   1    =    11

8    +   11   *   1    =    19

11  +   19   *   4    =    87

If you look at the sequence of numerators in the convergents of e ( 2, 3, 8, 11, … ) and the sequence of numbers in the convergents of ?2 ( 1, 2, 1, 1, 4, 1, … ), given the current numerator ( n ) and the previous numerator ( n-1 ) in the sequence and the corresponding number in the convergents of ?2 ( i )the next numerator ( n+1 ) can be calculated using the formula:

( n+1) = ( n-1 ) + n * i

Once we have this formula to work with, the rest is simple, the solution runs in 7 milliseconds on my machine.

# Project Euler – Problem 71 Solution

#### Problem

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

It can be seen that 2/5 is the fraction immediately to the left of 3/7.

By listing the set of reduced proper fractions for d <= 1,000,000 in ascending order of size, find the numerator of the fraction immediately to the left of 3/7.

#### Solution

This problem is fairly easy, given that the answer we’re looking for much be very close to 3 / 7 (0.4285714286) I simply iterate through the denominators, d, and find the closest numerator, n, which will produce a value less than 3 / 7. Then filter the set so we end up with only the n, d pairs that have a GCD of 1 and pick the numerator from the n, d pair whose n / d fraction is the biggest.