This proble is similiar to Projet Euler 76.
The Python program:
''' a_1*x_1+a_2*x_2+...+a_V*V=N A=[a_1,a_2,...,a_V] A=[2,3,5,7,11,...] ''' try: import psyco psyco.full() except ImportError: pass from euler import allprime def P(V,N): if N< =3 or V==[2]: return 1 elif N<V[-1]: return P(V[:-1],N) elif V==[2,3]: return N/6+1 else: return sum([P(V[:-1],N-V[-1]*x) for x in range(N/V[-1]+1)]) V,N=sorted(allprime(100).keys()),1 while P(V,N)< 5e3: N=N+1 print N
The problem is equivalent to obtain the number of the integer solutions of the diophantine equation a_1*x_1+a_2*x_2+…+a_V*V=N.
We define the number of the integer solutions of this diophantine equation as P（V,N).
It can be found that P(V,N) can be expressed as a recursive function as the following:
P(V,N)=P(V-1,N)+P(V-1,N-V)+…+P(V-1,N-[N/V]*V).
So This problem can be easily solved by defining this recursive function.
''' The number of the integer solutions of a_1*x_1+a_2*x_2+...+a_V*V=N ''' try: import psyco psyco.full() except ImportError: pass def P(V,N): if N< =1 or V==2: return N/2+1 elif V==1: return 1 elif N<V: return P(N-1,N)+1 else: return sum([P(V-1,N-V*x) for x in xrange(N/V+1)]) print P(99,100)
It is well known that if the square root of a natural number is not an integer, then it is irrational. The decimal expansion of such square roots is infinite without any repeating pattern at all.
The square root of two is 1.41421356237309504880…, and the digital sum of the first one hundred decimal digits is 475.
For the first one hundred natural numbers, find the total of the digital sums of the first one hundred decimal digits for all the irrational square roots.
import math from sympy import sqrt,evalf a=filter(lambda x:math.sqrt(x)>int(math.sqrt(x)),range(1,100)) for i in a: k=list(str(sqrt(i).evalf(105))) k.remove('.') s=s+sum(map(int,k[:100])) print s
This question is from ProjectEuler 04.
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 
99.
Find the largest palindrome made from the product of two 3-digit numbers.
Answer:
def palindromic(n): s=list(str(n)) if s==s[::-1]: return 1 else: return 0 m,n=999,99 p=[] for i in range(m,n,-1): for j in range(m,i-1,-1): if palindromic(i*j): p.append(i*j) break print max(p)
from math import sqrt def isprime(n): if n==1: return 0 elif n==2 or n==3: return 1 elif n%2==0 or n%3==0 or sqrt(n)==int(sqrt(n)): return 0 a=1 N=int(sqrt(n))+2 for x in range(3,N,2): if n%x==0: a=0 break return a
def factor(n): result=[] while n%2==0: result.append(2) n=n/2 while n!=1: if isprime(n): result.append(n) break j=3 for i in range(j,int(sqrt(n))+1,2): if n%i==0: result.append(i) n=n/i j=i break return result