5, 3, 2, Go!

Computer Science Level 3

2017 2003 2001 … 7 5 3 2 \large { {2017 \, 2003 \, 2001 \ldots 7 \, 5 \, 3 \, 2} }

The above expression shows the concatenation of the prime numbers less than or equal to 2017 written backwards.

Let L L and S S denote the number of digits and the sum of digits, respectively, of this large number. What can you say about the number obtained from the expression given below? 1 10 [ ( S − L ) ( m o d 2016 ) ] \dfrac1{10} \left[ (S-L) \pmod{2016} \right]

This problem is a part of set Prime Crimes via Computer Science
It is a perfect cube It is a prime number It is a number less than 100 It is a perfect square

Zeeshan Ali by Zeeshan Ali , 25, Pakistan

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1 solution

David Holcer
Mar 17, 2016 
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def primes(n):
    """ Returns  a list of primes < n """
    sieve = [True] * n
    for i in xrange(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
    return [2] + [i for i in xrange(3,n,2) if sieve[i]]

def digsum(n):
    summ=0
    for i in str(n):
        summ+=int(i)
    return summ

conc=''
for i in primes(2018): conc+=str(i)
l=len(conc)
s=digsum(conc)
print ((s-l)%2016)/10

I used a primes under n generator to find all primes under 2017, added them to an empty string, and found l and s.

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