;; Euler published the remarkable quadratic formula:

;; It turns out that the formula will produce 40 primes for the

;; consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41

;; = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41

-;; 41˛ + 41 + 41 is clearly divisible by 41.

-;; Using computers, the incredible formula n˛ - 79n + 1601 was discovered

+;; 41Â˛ + 41 + 41 is clearly divisible by 41.

+;; Using computers, the incredible formula nÂ˛ - 79n + 1601 was discovered

;; which produces 80 primes for the consecutive values n = 0 to 79. The

;; product of the coefficients, -79 and 1601, is -126479.

;; Considering quadratics of the form:

-;; n˛ + an + b, where |a| 1000 and |b| 1000

+;; nÂ˛ + an + b, where |a| 1000 and |b| 1000

;; where |n| is the modulus/absolute value of n

;; e.g. |11| = 11 and |-4| = 4

;{{{ problem 31 -- currency

-;; In England the currency is made up of pound, Ł, and pence, p, and

+;; In England the currency is made up of pound, ÂŁ, and pence, p, and

;; there are eight coins in general circulation:

-;; 1p, 2p, 5p, 10p, 20p, 50p, Ł1 (100p) and Ł2 (200p).

+;; 1p, 2p, 5p, 10p, 20p, 50p, ÂŁ1 (100p) and ÂŁ2 (200p).

(def denominations '(200 100 50 20 10 5 2 1))

-;; It is possible to make Ł2 in the following way:

-;; 1Ł1 + 150p + 220p + 15p + 12p + 31p

-;; How many different ways can Ł2 be made using any number of coins?

+;; It is possible to make ÂŁ2 in the following way:

+;; 1ÂŁ1 + 150p + 220p + 15p + 12p + 31p

+;; How many different ways can ÂŁ2 be made using any number of coins?

(defn coins [value denominations]

;; (println value denominations (range (inc (/ value (first denominations)))))

;{{{ problem 42 -- triangle words

;; The nth term of the sequence of triangle numbers is given by, tn =

-;; ˝n(n+1); so the first ten triangle numbers are:

+;; Â˝n(n+1); so the first ten triangle numbers are:

;; 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

;; The first three consecutive numbers to have three distinct prime factors are: