jakedobkin

# http://projecteuler.net/problem=94
# we're going to solve this using diophantine equations 
# first convert area formula to -3a**2-2a+4c**2+1=0 and -3a**2+2a+4c**2+1=0
# then run that through this solver http://www.alpertron.com.ar/QUAD.HTM to get 2 recursive equations

limit = 1000000000
sum = 0

# a-1 case
a = 1

jakedobkin

# http://projecteuler.net/problem=100
# algabraic rearrangement to 2b**2-2b-t**2+t = 0 (where b is blue balls, t is total balls)
# this is a diophantine equation, which can be solved recursively- to get formulas for next b,t:
# http://www.alpertron.com.ar/QUAD.HTM, put in coeeficients above - got that from Dreamshire

b = 85
t = 120

while t < 10**12:
# i learned that this allows you to set both at once so you don't have to use an intermediate variable

jakedobkin

# http://projecteuler.net/problem=85
# notes- formula to count boxes is just x * x+1 / 2  * y * y+1 /2 
# this is triangle number x * triangle number y

def triangle(n):
	return n*(n+1)/2

# i found the ranges by trial and error, starting at 100 each
min = 2000000
for a in range (0,37):

jakedobkin

# http://projecteuler.net/problem=95
# brute force sieve
n=1000000
sequences = {}
divisorSum = [1]*(n+1)
for i in range (2,n/2+1):
	j = 2*i
	while j<=n:
		divisorSum[j]+=i
		j = j+i

jakedobkin

# http://projecteuler.net/problem=87

# set up prime sieve to 50MM**.25
n=7100
myPrimes = [True]*(n+1)
last = 0
for i in range (2,n):
	if myPrimes[i] == True:
		j = 2*i
		while j<=n:

jakedobkin

# this is the simplest way- show all but the last 10 digits
print str(28433*pow(2,7830457)+1)[2357197:]

# which is equivalent to 
print (28433 * 2**7830457 + 1) % 10000000000

# which is equivalent to
print (28433 * pow(2, 7830457) + 1) % 10**10

# which is equivalent to

jakedobkin

# http://projecteuler.net/problem=99

from math import log

file = open('base_exp.txt','r').readlines()

# i thought maybe if we worked the box in two sections- from the top down and from the bottom up
# meeting in the middle, where the target cell was- but this doesn't appear to work either
array = []
for i in range (0,len(file)):

jakedobkin

# http://projecteuler.net/problem=92
# this version takes advantage of the fact that all numbers up to 10MM have roundfadds < 600
# so first compute those, then check them for each number from 1 to 10MM
# that shaves it from 6 minutes to 120s 

def fadd(n):
	sum = 0
	for i in range (0,len(str(n))):
		sum += int(str(n)[i:i+1])**2
	return sum

jakedobkin

file = open('matrix2.txt','r').readlines()

# i thought maybe if we worked the box in two sections- from the top down and from the bottom up
# meeting in the middle, where the target cell was- but this doesn't appear to work either
array = []
for i in range (0,len(file)):
	line = file[i].split(',')
	array.append(line)
n = len(array)

jakedobkin

file = open('matrix.txt','r').readlines()

array = []
for i in range (0,len(file)):
	line = file[i].split(',')
	array.append(line)

#adding algorithm works backward from next to last row to 0 row
for r in range (79,-1,-1):
	for s in range (79,-1,-1):