Pentagonal numbers are generated by the formula, $P_n=\frac{1}{2}n(3n-1)$. The first ten pentagonal numbers are: \[1,5,12,22,35,51,70,92,117,145,\dots\]
It can be seen that $P_4+P_7=22+70=92=P_8$. However, their difference, $70-22=48$, is not pentagonal.
Find the pair of pentagonal numbers, $P_j$ and $P_k$, for which their sum and difference are pentagonal and $D=\lvert P_j - P_k\rvert$.
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.
The $n^{\text{th}}$ 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: \[1,3,6,10,15,21,28,36,45,55,\dots\]
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?
An irrational decimal fraction is created by concatenating the positive integers:
0.123456789101112131415161718192021...
It can be seen that the 12th digit of the fractional part is 1.
If $d_n$ represents the $n^{\text{th}}$ digit of the fractional part, find the value of the following expression: \[d_1\times d_{10}\times d_{100}\times d_{1000}\times d_{10000}\times d_{100000}\times d_{1000000}\]
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?
If $p$ is the perimeter of a right angle triangle with integral length sides, ${a,b,c}$, there are exactly three solutions for $p = 120$. \[\begin{aligned} \{20,48,52\},\{24,45,51\},\{30,40,50\} \end{aligned}\]
For which value of $p\leq1000$, is the number of solutions maximised?
Take the number 192 and multiply it by each of 1, 2, and 3: \[192\times1=192 \\ 192\times2=384 \\ 192\times3=576\]
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?
The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, 3.
Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
2, 3, 5, and 7 are not considered to be truncatable primes.