Consider the following "magic" 3-gon ring, filled with the numbers 1 to 6, and each line adding to nine.

Working clockwise, and starting from the group of three with the numerically lowest external node (4,3,2 in this example), each solution can be described uniquely. For example, the above solution can be described by the set: 4,3,2; 6,2,1; 5,1,3.

It is possible to complete the ring with four different totals: 9, 10, 11, and 12. There are eight solutions in total.

By concatenating each group it is possible to form 9-digit strings; the maximum string for a 3-gon ring is 432621513.

Using the numbers 1 to 10, and depending on arrangements, it is possible to form 16- and 17-digit strings. What is the maximum 16-digit string for a "magic" 5-gon ring?

Analysis

see Solution.

Solution

The clues “magic” and 16-digit not 17-digit number give this one away. Just as the example, the bigger numbers go on the outside and the smaller numbers on the inside. They total 14 from all views, and, following the instructions above:
Working clockwise, and starting from the group of three with the numerically lowest external node, each solution can be described uniquely.

On our first try we discovered the answer and have included it below, so don’t look if you want to face the challenge yourself.

Comments

The square root of 2 can be written as an infinite continued fraction.

√2 = 1 +	1
	2 +	1
		2 +	1
			2 +	1
				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.

1 +	1	= 3/2
	2

1 +	1		= 7/5
	2 +	1
		2

1 +	1			= 17/12
	2 +	1
		2 +	1
			2

1 +	1				= 41/29
	2 +	1
		2 +	1
			2 +	1
				2

Hence the sequence of the first ten convergents for √2 are:

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:

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

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

Analysis

Continue the series for the numerator to 100 terms and print the sum of the digits.

Solution

Runs < 1 second in Python.

n, d, i = 2, 1, 100
while i>0:
  c = 1
  if i%3 == 0: c = 2*i/3
  n, d, i = d, c*d+n, i-1
print "Answer to PE65 = ", sum(int(i) for i in str(d))

Comments

Euler’s Totient function, φ(n) [sometimes called the phi function], is used to determine the number of numbers less than 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.

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.

Analysis

We’re trying to find the smallest φ(n) and the largest n in order to keep n/φ(n) at a maximum. One method is to multiply as many consecutive primes as possible without exceeding the limit of 1,000,000.

Solution

Runs < 1 seconds in Python.

primes = [2,3,5,7,11,13,17,19,23,27,31,37,41]
limit=10**6
max = 1
for p in primes:
  if max * p > limit: break
  max *= p
print "Answer to PE69 = ", max

Comments

for 10⁷? 9,699,690 = 2 x 3 x 5 x 7 x 11 x 13 x 17 x 19, and following the same method for 10⁹ would be 223,092,870

The cube, 41063625 (345³), can be permuted to produce two other cubes: 56623104 (384³) and 66430125 (405³). In fact, 41063625 is the smallest cube which has exactly three permutations of its digits which are also cube.

Find the smallest cube for which exactly five permutations of its digits are cube.

Analysis

Well after establishing that 10,000 is as far as we need to go, we brute force a search by cubing all the counting numbers in the loop’s range and sorting the digits in ascending order to use as a hash key. Then we increment a counter, %sd, and assign the first cube to our %cubes hash using that key. This way we are able to, later after the loop completes, grep out those cubes that belong to a family of 5 and find the minimum.

Solution

my $limit = 10_000;
my $min_cube = 9e20;
 
for (my $x=100; $x<$limit; $x++, $c=$x**3) {
  $cs = join('', sort split //,$c);
  $sd{$cs}++;
  $cubes{$cs}||=$c;
}
$min_cube = min_n($min_cube, $cubes{$_}) for grep {$sd{$_}==5} keys %sd;
print "Answer to PE62 = $min_cube";

Comments

The Perl short circuit assignment operator, ||=, simply assigns a value to a variable only if that variable is undefined or zero. Just a way to make sure the lowest value is set (since the loop is ascending) and never replaced.
The series of cube roots are: 5027 7061 7202 8288 8384

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 5
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom in triangle.txt (right click and ‘Save Link/Target As…’), a 15K text file containing a triangle with one-hundred rows.

NOTE: This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are 2⁹⁹ altogether! If you could check one trillion (10¹²) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;o)

Analysis

Used the program from problem 18. It takes less than a second as compared to waiting for the sun to burn out.

Solution

Runs < 1 second in Perl.

my @rows = map { [split /s/] } <DATA>;
my $nrows = $#rows;
 
for (my $i = $nrows; $i>0; $i--) {
  for my $j (0 .. $i-1) { 
    $rows[$i-1][$j] += max_n( $rows[$i][$j], $rows[$i][$j+1] );
  }
}
print "Answer to PE67 = $rows[0][0]";
__DATA__
3
7 5
2 4 6
8 5 9 3

Comments

• We’ve included the sample data as an example.

The first line of the program reads the data file, which is appended to the end of the program after the __DATA__ separator, into a two dimensional array named @rows. $nrows is the last index of the array so in this case 3.

• Check here for more information on max_n().
• See problem 18, problem 81, 82 & 83.

The 5-digit number, 16807=7⁵, is also a fifth power. Similarly, the 9-digit number, 134217728=8⁹, is a ninth power.

How many n-digit positive integers exist which are also an nth power?

Analysis

Note: log() = log₁₀; ln() = log_e. One way to know how many digits a number has is to use int(log n) +1 or if you don’t have a log function then use the natural log function as int(ln n / ln 10) +1. For this problem n cannot be greater then 10 since 10ⁿ is always n+1 digit long.

So we need to create a function f such that f(n) = upper limit of n. Using our log function and using 9 as an example we define f(9) as int (1 / (1 – log₁₀ 9)) or 21 such that 9ⁿ < 10^n-1. Thus 9ⁿ is n digits long when n ≤ 21. We repeat this for 1 through 8 and sum for an answer.

Solution

Comments

We didn’t see a need to write a program for this one.