def f(fraction, i): #fraction = (num,den)
frac_num = fraction[0]
frac_den = fraction[1]
new_frac_num = frac_den
new_frac_den = numbers[i]*frac_den + frac_num
g = gcd(new_frac_den, new_frac_num)
new_frac_den /= g
new_frac_num /= g
if i ==0: return new_frac_den
return f((new_frac_num,new_frac_den),i-1)
def E69():
res = 0
max_res = 0
max_i = 0
for i in range (2, MAXNUM):
if i % 2 == 1:
continue;
if is_prime(i) == True:
continue
cnt = 1
for j in range (1, i):
if gcd(i, j) == 1:
cnt+=1
if cnt != 0:
res = float(i)/float(cnt)
if max_res < res:
max_res = res
max_i = i
print max_res, max_i
return 0
def E69():
res = 0
max_res = 0
max_i = 0
for i in range(2, MAXNUM):
if i % 2 == 1:
continue
if is_prime(i) == True:
continue
cnt = 1
for j in range(1, i):
if gcd(i, j) == 1:
cnt += 1
if cnt != 0:
res = float(i) / float(cnt)
if max_res < res:
max_res = res
max_i = i
print max_res, max_i
return 0
triangles2(i)
"""
asa=[]
for i in range(1500001):
asa.append(0)
t=[]
#620x620
#var result = myList.OrderBy(k => k.Item2);
limit=sqrt(1500000/float(2))
print int(limit)+1
for n in range(1,int(limit)+1):
for m in range(1,int(limit)+1):
if m>n and gcd(m,n)==1 and (m-n)%2!=0:
#print m*m-n*n,2*m*n,m*m+n*n,2*m*m+2*m*n
a=m*m-n*n
b=2*m*n
c=m*m+n*n
L=2*m*m+2*m*n
t.append((a,b,c,L))
for s in range(L,1500001,L):
asa[s]+=1
print t
"""
for i in range(0,len(t)):
for k in range(2,(t[len(t)-1][3]/t[i][3])+1):
t.append((t[i][0]*k,t[i][1]*k,t[i][2]*k,t[i][3]*k))
'''
This is just an extension of Problem 9 of project euler.
I just initiated the value, then took a squareroot( since the values req. are 3)
Then did a gcd b/w the 2 values to find the later value.
Then did a sum to get the Length. Then added them and printed them out.
'''
from Euler import gcd, sqrt
L = 1500000
sqrt_L = int(sqrt(L))
lengths = [0]*L #initating the arrays.
for i in range(1, sqrt_L, 2):
for j in range(2, sqrt_L- i , 2):
if gcd(i, j) == 1:
sum = abs(j*j - i*i) + 2*i*j + i*i + j*j
for s in range(sum, L, sum):
lengths[s]+=1
print lengths.count(1)
from math import floor, ceil
from Euler import gcd
def mainly_slow ():
frac_set = set()
#for d in range(2, 100+1):
for d in range(2, 12*10**3+1):
if not d%100:
print d
# for this d, get the ints that bound 1/3 & 2/3
n_min = int(ceil(d/3.))
n_max = int(floor(d/2.))
for n in range(n_min, n_max+1):
f = Fraction(n,d)
frac_set.add(f)
print '\nAnswer:', len(frac_set)-2
pass
total = 0
for d in range(3, 12*10**3+1):
for n in range(int(ceil(d/3.)), int(floor(d/2.))):
if gcd(n,d) == 1:
total += 1
print total
"""
This is just an extension of Problem 9 of project euler.
I just initiated the value, then took a squareroot( since the values req. are 3)
Then did a gcd b/w the 2 values to find the later value.
Then did a sum to get the Length. Then added them and printed them out.
"""
from Euler import gcd, sqrt
L = 1500000
sqrt_L = int(sqrt(L))
lengths = [0] * L # initating the arrays.
for i in range(1, sqrt_L, 2):
for j in range(2, sqrt_L - i, 2):
if gcd(i, j) == 1:
sum = abs(j * j - i * i) + 2 * i * j + i * i + j * j
for s in range(sum, L, sum):
lengths[s] += 1
print lengths.count(1)
from Euler import gcd
L = 12000+1
c = 0
for n in range(5, L):
for k in range(n/3 + 1, (n-1)/2 + 1):
if gcd(n,k) == 1: c+=1
print "Answer to PE73 =", c
#91. Count the number of lattice point right triangles which use the origin
# in the 50x50 grid.
from Euler import gcd
count = 3*50**2 #base has slope 0
for x in xrange(1,51):
for y in xrange(1,51): #this will run over all the points (x,y) which
g = gcd(x,y) #form the right base point
x_new =x/g
y_new =y/g
k=1
while k <= x//y_new:
if y + k*x_new <=50:
count += 2
k += 1
print count
<div id="content">
<div style="text-align:center;" class="print"><img src="images/print_page_logo.png" alt="projecteuler.net" style="border:none;" /></div>
<h2>Counting fractions in a range</h2><div id="problem_info" class="info"><h3>Problem 73</h3><span>Published on Friday, 2nd July 2004, 06:00 pm; Solved by 15547; Difficulty rating: 15%</span></div>
<div class="problem_content" role="problem">
<p>Consider the fraction, <i>n/d</i>, where <i>n</i> and <i>d</i> are positive integers. If <i>n</i><<i>d</i> and HCF(<i>n,d</i>)=1, it is called a reduced proper fraction.</p>
<p>If we list the set of reduced proper fractions for <i>d</i> ≤ 8 in ascending order of size, we get:</p>
<p style="text-align:center;font-size:90%;">1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, <b>3/8, 2/5, 3/7</b>, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8</p>
<p>It can be seen that there are 3 fractions between 1/3 and 1/2.</p>
<p>How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for <i>d</i> ≤ 12,000?</p>
</div><br />
<br /></div>
"""
from Euler import gcd
MAXNUM=51
MAXNUM=12001
a2=1
a3=1
cnt = 0
for i in range(4, MAXNUM):
a2 = i*3
a3 = i*2
for j in range(a3+1, a2):
if j%6 == 0:
if gcd(i, j/6) == 1:
cnt+=1
print cnt