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 4
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.

This is a much more difficult version #18 - Maximum path sum I. It is not possible to try every route to solve this problem, as there are $2^{99}$ altogether! If you could check one trillion $\left(10^{12}\right)$ routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it ;o)

Each character on a computer is assigned a unique code and the preferred standard is ASCII (American Standard Code for Information Interchange). For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.

A modern encryption method is to take a text file, convert the bytes to ASCII, then XOR each byte with a given value, taken from a secret key. The advantage with the XOR function is that using the same encryption key on the cipher text, restores the plain text; for example, 65 XOR 42 = 107, then 107 XOR 42 = 65.

For unbreakable encryption, the key is the same length as the plain text message, and the key is made up of random bytes. The user would keep the encrypted message and the encryption key in different locations, and without both “halves”, it is impossible to decrypt the message.

Unfortunately, this method is impractical for most users, so the modified method is to use a password as a key. If the password is shorter than the message, which is likely, the key is repeated cyclically throughout the message. The balance for this method is using a sufficiently long password key for security, but short enough to be memorable.

Your task has been made easy, as the encryption key consists of three lower case characters. Using p059_cipher.txt (right click and ‘Save Link/Target As…’), a file containing the encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the ASCII values in the original text.

Starting with 1 and spiral anticlockwise in the following way, a square spiral with side length 7 is formed.

37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18  5  4  3 12 29
40 19  6  1  2 11 28
41 20  7  8  9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49

It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ~ 62%.

If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?

Is it possible to show that the square root of two can be expressed as an infinite continued fraction. \[\sqrt{2} = 1 + \frac{1}{2+\frac{1}{2 + \frac{1}{2+\dots}}} = 1.414213\dots\]

By expanding this for the first four iterations, we get: \[\begin{aligned} 1 + \frac{1}{2} &= \frac{3}{2} = 1.5 \\ 1 + \frac{1}{2+\frac{1}{2}} &= \frac{7}{5} = 1.4 \\ 1 + \frac{1}{2+\frac{1}{2+\frac{1}{2}}} &= \frac{17}{12} = 1.41666\dots \\ 1 + \frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}} &= \frac{41}{29} = 1.41379\dots \end{aligned}\]

The next three expansions are, $\frac{99}{70},\frac{239}{169},\frac{577}{408}$, but the eight expansion, $\frac{1393}{895}$, 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?

A googol $\left(10^{100}\right)$ is a massive number: one followed by one-hundred zeros; $100^{100}$ is almost unimaginably large: one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is only 1.

Considering natural numbers of the form, $a^b$, where $a, b < 100$, what is the maximum digital sum?

If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.

Not all numbers produce palindromes so quickly. For example, \[\begin{aligned} 349+943 &= 1292 \\ 1292+2921 &= 4213 \\ 4213+3124 &= 7337 \end{aligned}\]

That is, 349 took three iterations to arrive at a palindrome.

Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).

Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is 4994.

How many Lychrel numbers are there below ten-thousand?

It can be seen that the number, 125874, and its double 251748, contain exactly the same digits, but in a different order.

Find the smallest positive integer, $x$, such that $2x,3x,4x,5x$ and $6x$, contain the same digits.

For a number written in Roman numerals to be considered valid there are basic rules which must be followed. Even though the rules allow some numbers to be expressed in more than one way there is always a “best” way of writing a particular number.

For example, it would appear that there are at least six ways of writing the number sixteen:

IIIIIIIIIIIIIIII
VIIIIIIIIIII
VVIIIIII
XIIIIII
VVVI
XVI

However, according to the rules only XIIIIII and XVI are valid, and the last example is considered to be the most efficient, as it uses the least number of numerals.

The 11K text file, roman.txt (right click and ‘Save Link/Target As…’), contains one thousand numbers written in valid, but not necessarily minimal, Roman numerals; see About… Roman Numerals for the definitive rules for this problem.

Find the number of characters saved by writing each of these in their minimal form.

Note: You can assume that all the Roman numerals in the file contain no more than four consecutive identical units.

A common security method used for online banking is to ask the user for three random characters from a passcode. For example, if the passcode was 531278, they may ask for the 2nd, 3rd, and 5th characters; the expected reply would be: 317.

The text file, keylog.txt, contains fifty successful login attempts.

Given that the three characters are always asked for in order, analyze the file so as to determine the shortest possible secret passcode of unknown length.

The square root of 2 can be written as an infinite continued fraction. \[\sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2+\dots}}}}\]

The infinite continued fraction can be written, $\sqrt{2}=[1;(2)]$, $(2)$ indicates that 2 repeats ad infinitum. In a similar way, $\sqrt{23}=[4;(1,3,1,8)]$.

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 $\sqrt{2}$. \[\begin{aligned} 1 + \frac{1}{2} &= \frac{3}{2} \\ 1 + \frac{1}{2 + \frac{1}{2}} &= \frac{7}{5} \\ 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2}}} &= \frac{17}{12} \\ 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2}}}} &= \frac{41}{29} \end{aligned}\]

Hence the sequence of the first ten convergents for $\sqrt{2}$ are: \[1,\frac{3}{2}, \frac{7}{5}, \frac{17}{12}, \frac{41}{29}, \frac{99}{70}, \frac{239}{169}, \frac{577}{408}, \frac{1393}{985}, \frac{3363}{2378}, \dots\]

What is most surprising is that the important mathematical constant, \[e=[2;1,2,1,1,4,1,1,6,1,\dots,1,2k,1,\dots]\]

The first ten terms in the sequence of convergents for $e$ are: \[2,3,\frac{8}{3},\frac{11}{4},\frac{19}{7}, \frac{87}{32}, \frac{106}{39}, \frac{193}{71}, \frac{1264}{465}, \frac{1457}{536},\dots\]

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

Find the sum of digits in the numerator of the $100^\text{th}$ convergent of the continued fraction for $e$.