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.
Consider all integer combinations of $a^b$ for $2\leq a\leq 5$ and $2\leq b\leq 5$:
Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
21 22 23 24 25 20 7 8 9 10 19 6 1 2 11 18 5 4 3 12 17 16 15 14 13It can be verified that the sum of the numbers on the diagonals is 101.
What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?
Euler discovered the remarkable quadratic formula: \[n^2+n+41\]
It turns out that the formula will produce 40 primes for the consecutive integer values $0\leq n\leq 39$. However, when $n = 40, 40^2+40+41=40(40+1)+41$ is divisible by 41, and certainly when $n=41,41^2+41+41$ is clearly divisible by 41.
The incredible formula $n^2-79n+1601$ was discovered, which produces 80 primes for the consecutive values $0\leq n\leq 79$. The product of the coefficients, -79 and 1601, is -126479.
Considering quadratics of the form: \[n^2+an+b,\text{ where } |a|<1000\text{ and } |b|\leq 1000\]
where the bars indicate absolute value.
Find the product of the coefficients, $a$ and $b$, for the quadratic expression that produces the maximum number of primes for consecutive values of $n$, starting $n=0$.
A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given: \[\begin{aligned} \frac{1}{2} &= 0.5 \\ \frac{1}{3} &= 0.\overline{3} \\ \frac{1}{4} &= 0.25 \\ \frac{1}{5} &= 0.2 \\ \frac{1}{6} &= 0.1\overline{6} \\ \frac{1}{7} &= 0.\overline{142857} \\ \frac{1}{8} &= 0.125 \\ \frac{1}{9} &= 0.\overline{1} \\ \frac{1}{10} &= 0.1 \end{aligned}\]
where $0.1\overline{6}$ means 0.166666…, and has a 1-digit recurring cycle. It can be seen that $\frac{1}{7}$ has a 6-digit recurring cycle.
Find the value of $d<1000$ for which $\frac{1}{d}$ contains the longest recurring cycle in its decimal fraction part.
