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.

The decimal number, $585=1001001001_2$ (binary), is palindromic in both bases.

Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.

(Please note that the palindromic number, in either base, may not include leading zeros.)

The number, 197, is called a circular prime because all rotations of the digits: 197, 971, 719, are themselves prime.

There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.

How many circular primes are there below one million?

145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.

Find the sum of all numbers which are equal to the sum of the factorial of their digits.

As 1! = 1 and 2! = 2 are not sums they are not included.

The fraction $\frac{49}{98}$ is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that $\frac{49}{98} = \frac{4}{8}$, which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, $\frac{30}{50} = \frac{3}{5}$, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.

If the product of these four fractions is given in its lowest common terms, find the value of the denominator.

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, the 5-digit number, 15234, is 1 through 5 pandigital.

The product 7254 is unusual, as the identity, $39\times 186=7254$, containing multiplicand, multiplier, and product is 1 through 9 pandigital.

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

Some products can be obtained in more than one way so be sure to only include it once in your sum.

In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:

1p, 2p, 5p, 10p, 20p, 50p, £1 (100p), £2 (200p)

It is possible to make £2 in the following way:

1x£1 + 1x50p + 2x20p + 1x5p + 1x2p + 3x1p

How many different ways can £2 be made using any number of coins?

Surprisingly there are only three numbers that can be written as the sum of the fourth powers of their digits: \[\begin{aligned} 1634 &= 1^4 + 6^4 + 3^4 + 4^4 \\ 8208 &= 8^4 + 2^4 + 0^4 + 8^4 \\ 9474 &= 9^4 + 4^4 + 7^4 + 4^4 \end{aligned}\]

As $1=1^4$ is not a sum it is not included.

The sum of these numbers is 1634 + 8208 + 9474 = 19316.

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.