Apocalyptic numbers

Computer Science Level 3

Find the least n n such that the number 2 n 2^n contain 3 consecutive 6's, " 666 666 ".


The answer is 157.

Abdeslem Smahi by Abdeslem Smahi , 24, Algeria

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4 solutions

Arulx Z
Jan 23, 2016 
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>>> n = 2
>>> count = 1
>>> while '666' not in str(n):
        n *= 2
        count += 1

>>> print count
157

2 Helpful 0 Interesting 0 Brilliant 0 Confused

i really like your solutions ^^

Abdeslem Smahi - 5 years, 4 months ago

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Thanks a lot :)

Arulx Z - 5 years, 4 months ago
David Holcer
Mar 17, 2016 
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c=0
while True:
    num=str(2**c)
    if '666' in num:
        print c
        break
    c+=1

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Bill Bell
Jan 23, 2016

Avoiding the use of strings and of exponentiation, both of which are expensive. 

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def threeConsecutive(k,targetDigit=6,targetCount=3):
    found=0
    while k:
        digit,k=k%10,k/10
        if digit==targetDigit:
            found+=1
        else:
            found=0
        if found==targetCount:
            return True
    return False

n=1
consider=2
while True:
    consider*=2
    n+=1
    if threeConsecutive(consider):
        print n, consider
        break

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I do agree that string conversion is expensive but for weird reasons, it's actually faster.

Arulx Z - 5 years, 4 months ago

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Ah, very interesting. I must try timing it sometime.

Bill Bell - 5 years, 4 months ago

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Here are the results - 

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1000 loops, best of 3: 977 usec per loop


1
 
10000 loops, best of 3: 46.4 usec per loop

Arulx Z - 5 years, 4 months ago

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@Arulx Z – Here are some of my timings too. 

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>pythonw -u "time strings.py"
157 182687704666362864775460604089535377456991567872
timing: 0.00600004
157 182687704666362864775460604089535377456991567872
timing: 0.00100017
>Exit code: 0
>pythonw -u "time strings.py"
157 182687704666362864775460604089535377456991567872
timing: 0.00500011
157 182687704666362864775460604089535377456991567872
timing: 0.00099993
>Exit code: 0
>pythonw -u "time strings.py"
157 182687704666362864775460604089535377456991567872
timing: 0.00499988
157 182687704666362864775460604089535377456991567872
timing: 0.00099993
>Exit code: 0

In each case the first timing is for my original code, the second using equivalent string code. You're absolutely right. I'm old enough (69 in less than a week's time) that I should know never to depend on my intuitions.

Bill Bell - 5 years, 4 months ago

Here is my code (Python 3): 

def iscons(n):
    for i in range(0,len(str(n))-2):
        if int(str(n)[i])==6 and int(str(n)[i+1])==6 and int(str(n)[i+2])==6:
            return True
        else:
            pass
    return False

n=1
while n>0:
    if iscons(2**n)==True:
        print (n)
        break
    else:
        n+=1
0 Helpful 0 Interesting 0 Brilliant 0 Confused

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