Geek Answers Handbook

(first ten-digit number in e) .com python google challenge 2004

I just approached one of the problems allegedly used by Google 2004

(the first 10-digit prime in e).com

Regardless, I wanted to take the challenge and solve it using python

>>> '%0.52f' % math.exp(1)
'2.71828182845904509079**5598298427**6488423347473144531250'
>>> '%0.52f' % numpy.exp(1)
'2.71828182845904509079**5598298427**6488423347473144531250'

my program returned 5598298427which is a prime

after browsing the internet the correct answer was 7427466391

but the exp number in python does not include the numbers you see above

import numpy
import math

def prime(a):
    if a == 2: return True
    if a % 2 == 0: return False
    if a < 2: return False
    i = 2
    n = math.sqrt(a) + 1
    while(i < n):
        if a % i == 0:
            return False
        i += 1
    return True

def prime_e():
    e = '%0.51f' % math.exp(1)
    e = e.replace("2.","")
    for i in range(len(e)):
        x = int(e[i:10+i])
        if prime(x):
            return [i, x]

print prime_e()

Am I doing something wrong?

EDIT: using gmpy2

def exp():
    with gmpy2.local_context(gmpy2.context(), precision=100) as ctx:
        ctx.precision += 1000
        return gmpy2.exp(1)

returns 7427466391after 99 iterations

+4
source share
3 answers

Actual value of e (Euler constant)

http://www.gutenberg.org/files/127/127.txt

2,71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642 7427466391 932003059921817413596629043572900334295260595630...

- 7427466391. e math.exp(1)

+3

:

1 000 e, @quantum e one Digit , @wnoise 2, " Haskell ..., ":

def z(contfrac, a=1, b=0, c=0, d=1):
    for x in contfrac:
        while a > 0 and b > 0 and c > 0 and d > 0:
            t = a // c
            t2 = b // d
            if not t == t2:
                break
            yield t
            a = (10 * (a - c*t))
            b = (10 * (b - d*t))
            # continue with same fraction, don't pull new x
        a, b = x*a+b, a
        c, d = x*c+d, c
    for digit in rdigits(a, c):
        yield digit

def rdigits(p, q):
    while p > 0:
        if p > q:
           d = p // q
           p = p - q * d
        else:
           d = (10 * p) // q
           p = 10 * p - q * d
        yield d    

def e_cf_expansion():
    yield 1
    k = 0
    while True:
        yield k
        k += 2
        yield 1
        yield 1

def e_dec():
    return z(e_cf_expansion())

gen = e_dec()
e = [str(gen.next()) for i in xrange(1000)]
e.insert(1, '.')

, Rosetta Code Primality_by_trial_division # Python:

def isprime(a):
    if a < 2: return False
    if a == 2 or a == 3: return True # manually test 2 and 3   
    if a % 2 == 0 or a % 3 == 0: return False # exclude multiples of 2 and 3
    maxDivisor = a**0.5
    d, i = 5, 2
    while d <= maxDivisor:
        if a % d == 0: return False
        d += i 
        i = 6 - i # this modifies 2 into 4 and viceversa
    return True

10- e ( ):

for i in range(len(e[2:])-10):
  x = int(reduce(operator.add,e[2:][i:i+10]))
  if isprime(x):
      print x
      print i
      break

:

7427466391
98

, 10- e 98- , http://explorepdx.com/firsten.html "The ".

e , , Euler Number 100 (Python), Python Decimal :

import operator
import decimal as dc

def edigits(p):
    dc.getcontext().prec = p
    factorial = 1
    euler = 2
    for x in range(2, 150):
        factorial *= x
        euler += dc.Decimal(str(1.0))/dc.Decimal(str(factorial))
    return euler

estring = edigits(150).to_eng_string()[2:]

for i in range(len(estring)-10):
    x = int(reduce(operator.add,estring[i:i+10]))
    if isprime(x):
        print x
        print i
        break

:

7427466391
98

@MarkDickinson, - e . :

import operator
import decimal

decimal.getcontext().prec = 150
e_from_decimal = decimal.Decimal(1).exp().to_eng_string()[2:]
for i in range(len(e_from_decimal)-10):
    x = int(reduce(operator.add,e_from_decimal[i:i+10]))
    if isprime(x):
        print x
        print i
        break

:

7427466391
98
+1

, "e" 15- (09079 ) , . python , "e" . , . "" int, , . float int, 10, ( ) e_as_int = e * 10 ** d, d - . e :

import itertools
count = itertools.count

def ape(digits):
    # e = sum(1/k! for k in count())
    # calculate some extra digits to compensate for loss of precision:
    r = 3    
    m = 10**(digits+r)
    e = 2*m
    f = 1   #initial value for k!
    for k in count(2):
        f *= k
        if f>m: break
        # remember, we're doing int division, so m/f = 0 for f>m
        e += (m/f)

    return e/10**r #truncate to required precision

'ape' " e" :-) 10000 e .

0
source

More articles:


All Articles