Not a "Coincidence" Again!

Number Theory Level 5

$0. \,\ 029 \,\ 031 \,\ 037 \,\ 047 \,\ 061 \,\ 079 \,\ 101 \,\ 127 \,\ 157 \dots$

Above shows the first few digits decimal expansion of $\large{\frac{28944031}{997002999}}$ , written in groups of 3.

If we split this decimal expansion into groups of 3 we see multiple primes appearing! However these primes end at a certain point. After how many decimal (digits) does this pattern cease?

Details and Assumptions :

If you think the number equals to $0.\ 029 \,\ 031 \,\ 037 \,\ 047 \,\ 063$ , then you can only find the five prime numbers. So we can only find 4 positive prime numbers before the pattern breaks off. Thus the pattern breaks off after 12 decimal digits.

Inspired by Garret C. , Pi Han Goh and Daniel Liu .

The answer is 66.

2 solutions

Josh Speckman
Jul 11, 2015

Clearly the intended answer is bash away and hope it's not like 300. So we bash

The given fraction is equivalent to approximately $0.029031037047061079101127157191229271317367421479541607677751829911 998$ plus some stuff. We chunk this into triplets, and eventually get the triplet $998$ , which is clearly not prime. Thus, we just have to check all of the numbers before it. We find that none of the previous numbers are composite, so $998$ is the first composite chunk, and we count to get $66$ .

Seriously, though, what is the intended solution? I can't think of anything that feels useful.

David Peterson
Sep 3, 2015

n=1;

while isprime(floor(28944031/997002999 10^(3n)) - floor(28944031/997002999 10^(3 (n-1))) 10^(3))

n++;

cout << 3*n;

Not a "Coincidence" Again!

The answer is 66.

2 solutions

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