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 + 150p + 220p + 15p + 12p + 31p
How many different ways can £2 be made using any number of coins?

Problem Euler 31 - Coin sums
Using MSR Z3 SMT solver for solving this is overkill, and also slow, but nevertheless, it works, showing all possible solutions as well.
The piece of code for blocking already found solution and search for next, and thus, counting all solutions, was taken from here.

Constraints like "a>=0" should be present, because Z3 solver will search for solutions with negative numbers.

#!/usr/bin/python
from z3 import *
a,b,c,d,e,f,g,h = Ints('a b c d e f g h')
s = Solver()
s.add(1*a + 2*b + 5*c + 10*d + 20*e + 50*f + 100*g + 200*h == 200,
a>=0, b>=0, c>=0, d>=0, e>=0, f>=0, g>=0, h>=0)
result=[]
while True:
if s.check() == sat:
m = s.model()
print m
result.append(m)
# Create a new constraint the blocks the current model
block = []
for d in m:
# d is a declaration
if d.arity() > 0:
raise Z3Exception("uninterpreted functions are not suppported")
# create a constant from declaration
c=d()
#print c, m[d]
if is_array(c) or c.sort().kind() == Z3_UNINTERPRETED_SORT:
raise Z3Exception("arrays and uninterpreted sorts are not supported")
block.append(c != m[d])
#print "new constraint:",block
s.add(Or(block))
else:
print len(result)
break

Works very slow, and this is what it produces:

[h = 0, g = 0, f = 0, e = 0, d = 0, c = 0, b = 0, a = 200]
[f = 1, b = 5, a = 0, d = 1, g = 1, h = 0, c = 2, e = 1]
[f = 0, b = 1, a = 153, d = 0, g = 0, h = 0, c = 1, e = 2]
...
[f = 0, b = 31, a = 33, d = 2, g = 0, h = 0, c = 17, e = 0]
[f = 0, b = 30, a = 35, d = 2, g = 0, h = 0, c = 17, e = 0]
[f = 0, b = 5, a = 50, d = 2, g = 0, h = 0, c = 24, e = 0]

73682 results in total.