# Euler Problem 31¶

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) and £2 (200p).

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

1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p

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

Let's do it non-recursive. Tricky part is to use ceil, otherwise we get wrong ranges for example for $\frac{50}{20}$.

In [1]:

    from math import ceil

    print(list(range(50 // 20)))
    print(list(range(ceil(50 / 20))))

    [0, 1]
    [0, 1, 2]

In [2]:

    from math import ceil

    r = 200
    c = 0

    if r % 200 == 0:
        c += 1
    for p200 in range(ceil(r / 200)):
        r200 = r - p200 * 200
        if r200 % 100 == 0:
            c += 1
        for p100 in range(ceil(r200 / 100)):
            r100 = r200 - p100 * 100
            if r100 % 50 == 0:
                c += 1
            for p50 in range(ceil(r100 / 50)):
                r50 = r100 - p50 * 50
                if r50 % 20 == 0:
                    c += 1
                for p20 in range(ceil(r50 / 20)):
                    r20 = r50 - p20 * 20
                    if r20 <= 0:
                        break
                    if r20 % 10 == 0:
                        c += 1
                    for p10 in range(ceil(r20 / 10)):
                        r10 = r20 - p10 * 10
                        if r10 % 5 == 0:
                            c += 1
                        for p5 in range(ceil(r10 / 5)):
                            r5 = r10 - p5 * 5
                            if r5 % 2 == 0:
                                c += 1
                            for p2 in range(ceil(r5 / 2)):
                                r2 = r5 - p2 * 2
                                if r2 % 1 == 0:
                                    c += 1

    s = c

In [3]:

    print(s)
    assert(s == 73682)

    73682