Example #1
0
def pe(M):
    s = 0
    for m in range(2, isqrt(M/2)):
        mm = (M-1) // (2*m)
        m2 = m * m
        start = 1 if m % 2 == 0 else 2
        for n in range(start, min(mm-m+1, m), 2):
            if gcd(m, n) != 1:
                continue
            n2 = n * n
            if (m2+n2) % (m2-n2-2*m*n) != 0:
                continue
            s += mm // (m+n)
    return s
Example #2
0
def pe(M):
    s = 0
    for m in range(2, isqrt(M / 2)):
        mm = (M - 1) // (2 * m)
        m2 = m * m
        start = 1 if m % 2 == 0 else 2
        for n in range(start, min(mm - m + 1, m), 2):
            if gcd(m, n) != 1:
                continue
            n2 = n * n
            if (m2 + n2) % (m2 - n2 - 2 * m * n) != 0:
                continue
            s += mm // (m + n)
    return s
Example #3
0
def pe(M):
    rad = [0] * M
    rad[1] = 1
    for p in range(2, M):
        if rad[p] > 0: continue
        isqrtp = isqrt(p)
        flag = True
        for q in range(2, isqrtp + 1):
            if p % q != 0: continue
            pp = p
            while pp % q == 0:
                pp //= q
            qq = q * rad[pp]
            pp = p
            while pp < M:
                rad[pp] = qq
                pp *= q
            flag = (q > isqrtp)
            break
        if flag:
            pp = p
            while pp < M:
                rad[pp] = p
                pp *= p

    s = 0
    for c in range(3, M):
        cc = (c - 1) // rad[c]
        if rad[c - 1] <= cc:
            s += c
        if cc < 6: continue
        if c % 2 == 0 and cc < 15: continue
        if c % 3 == 0 and cc < 10: continue
        for a in range(2, c // 2):
            b = c - a
            if rad[a] > cc or rad[b] > cc: continue
            if rad[a] * rad[b] <= cc and gcd(a, b) == 1:
                s += c
    return s
Example #4
0
def pe(M):
    rad = [0] * M
    rad[1] = 1
    for p in range(2, M):
        if rad[p] > 0: continue
        isqrtp = isqrt(p)
        flag = True
        for q in range(2, isqrtp + 1):
            if p % q != 0: continue
            pp = p
            while pp % q == 0:
                pp //= q
            qq = q * rad[pp]
            pp = p
            while pp < M:
                rad[pp] = qq
                pp *= q
            flag = (q > isqrtp)
            break
        if flag:
            pp = p
            while pp < M:
                rad[pp] = p
                pp *= p

    s = 0
    for c in range(3, M):
        cc = (c-1)//rad[c]
        if rad[c-1] <= cc:
            s += c
        if cc < 6: continue
        if c % 2 == 0 and cc < 15: continue
        if c % 3 == 0 and cc < 10: continue
        for a in range(2, c//2):
            b = c - a
            if rad[a] > cc or rad[b] > cc: continue
            if rad[a] * rad[b] <= cc and gcd(a, b) == 1:
                s += c
    return s
Example #5
0
#Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can exactly one integer sided right angle triangle be formed?

#Note: This problem has been changed recently, please check that you are using the right parameters.

#Answer:
	#161667

from time import time; t=time()
from mathplus import gcd, isqrt

M = 1500000
q = [0]*(M+1)
for m in range(2, isqrt(M/2)+1):
    for n in range(m % 2 + 1, min(m, M//(2*m)-m+1), 2):
        #length = 2*m*(m+n)
        #if length > M: break
        #x = m*m-n*n
        #y = 2*m*n
        #if gcd(x,y)==1:
        if gcd(m, n)==1:
            length = 2*m*(m+n)
            q[length] += 1

p = [0]*(M+1)
for m, c in enumerate(q):
    if c:
        for n in range(m, M+1, m):
            p[n] += c

print(sum(1 for i in p if i == 1))#, time()-t
Example #6
0
#!/usr/bin/python
# -*- coding: utf-8 -*-

#Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.

#If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:

#1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8

#It can be seen that there are 3 fractions between 1/3 and 1/2.

#How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for d ≤ 12,000?

#Note: The upper limit has been changed recently.

#Answer:
	#7295372

from time import time; t=time()
from mathplus import gcd

D=12000

cnt = 0
for n in range(5, D+1):
    for m in range(n//3+1, (n+1)//2):
        if gcd(n, m) == 1: cnt += 1

print(cnt)#, time()-t
Example #7
0
t = time()
from mathplus import isqrt, gcd

M = 1000

p = [0] * (M // 2 + 1)
q = [0] * (M // 2 + 1)
for n in range(1, M // 2 + 1):
    for k in range(1, n):
        if n % k != 0: continue
        m = n // k
        if q[m] == 0:
            for i in range(isqrt(m // 2) + 1, isqrt(m) + 1):
                if m % i != 0: continue
                u, v = i, m // i - i
                if (u + v) % 2 != 0 and gcd(u, v) == 1:
                    q[m] += 1
                    #if n == 60:
                    #print k, u, v, k*(u*u-v*v), 2*k*u*v, k*(u*u+v*v)
        p[n] += q[m]

print(max((x, i) for i, x in enumerate(p))[1] * 2)  #, time()-t
'''
from time import time; t=time()
from mathplus import isqrt, gcd

M = 1000

ss = 0
MM = M/2+1
p = [0] * MM
Example #8
0
from time import time; t=time()
from mathplus import isqrt, gcd

M = 1000

p = [0] * (M//2+1)
q = [0] * (M//2+1)
for n in range(1, M//2+1):
    for k in range(1, n):
        if n % k != 0: continue
        m = n//k
        if q[m] == 0:
            for i in range(isqrt(m//2)+1, isqrt(m)+1):
                if m % i != 0: continue
                u, v = i, m//i - i
                if (u+v) % 2 != 0 and gcd(u, v) == 1:
                    q[m] += 1
                    #if n == 60: 
                        #print k, u, v, k*(u*u-v*v), 2*k*u*v, k*(u*u+v*v)
        p[n] += q[m]

print(max((x, i) for i, x in enumerate(p))[1]*2)#, time()-t
'''
from time import time; t=time()
from mathplus import isqrt, gcd

M = 1000

ss = 0
MM = M/2+1
p = [0] * MM