Problem 49(Prime permutations)

Problem 49(Prime permutations)

The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.

There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.

What 12-digit number do you form by concatenating the three terms in this sequence?

 

In Python:

import time
import math

def seive(n):
    i = 2
    primecodeList =[1 for i in range(1, n + 1)]
    primecodeList[0] = 0
    primeList = []
    while i < n + 1:
        if primecodeList[i - 1] == 1:
            primeList.append(i)
            for j in range(2, n // i + 1):
                primecodeList[i * j - 1] = 0
        i += 1
    return primeList

def ispermutation(i, j, k):
    stri = sorted(str(i))
    strj = sorted(str(j))
    strk = sorted(str(k))
    if stri == strj and strj == strk:
        return True

startTime = time.time()
primeList = []
for prime in seive(10 ** 4):
    if math.log10(prime) > 3:
        primeList.append(prime)

for i in range(0, len(primeList)):
    for k in range(0, i):
        j = (primeList[i] + primeList[k]) // 2
        if j in primeList and ispermutation(primeList[i], j, primeList[k]):
            print(str(primeList[k]) + str(j) + str(primeList[i]))
print(time.time() - startTime, "seconds")

Run time: 8.202334880828857 seconds

 

 

In Python:

import time
import math

def seive(n):
    i = 2
    primecodeList =[1 for i in range(1, n + 1)]
    primecodeList[0] = 0
    primeList = []
    while i < n + 1:
        if primecodeList[i - 1] == 1:
            for j in range(2, n // i + 1):
                primecodeList[i * j - 1] = 0
            if math.log10(i) > 3:
                primeList.append(i)
        i += 1
    return primeList

def ispermutation(i, j):
    stri = sorted(str(i))
    strj = sorted(str(j))
    if stri == strj:
        return True

startTime = time.time()
primeList = seive(10 ** 4)
for i in range(0, len(primeList)):
    for k in range(0, i):
        if ispermutation(primeList[i], primeList[k]):
            j = (primeList[i] + primeList[k]) // 2
            if j in primeList and ispermutation(primeList[i], j):
                print(str(primeList[k]) + str(j) + str(primeList[i]))
print(time.time() - startTime, "seconds")

Run time: 0.8857400417327881 seconds

 

 

Solution: 296962999629