Math brains: What are the 1st 3 prime numbers, without skipping over a prime, that each end in 7?

Question

Here's an answer - but not the first group:

4,297
4,327
4,337

These prime numbers are all in order with no other prime numbers between them.

Answer 1

1627 1637 1657

I'm wondering if there is an analytic way to solve this ??

Answer 2

People have already posted the answer. I can't think of a real clear short cut, but I've thought of few ways to narrow down the search.

First of all, the smallest of the three and the biggest of the three have to have a difference of 30 or more. This is because if the smallest of these is p, and is 1 mod 3, then the next number that ends in 7 is p+10 which is 2 mod 3, and then p+20 would be a multiple of 3. Likewise p+30 would be 1 mod 3 and the cycle continues again.

Assume (and this is a bit of a wild assumption) that two of the primes we want do in fact have a difference of exactly 30 like this. This also means that there has to be one multiple of 30 within this range. Since p-1 has to be a multiple of 3, then A multiple of 30 has to end in 0, this means p+23 (the only number in the range that ends in 0 and is also a multiple of 23) is the multiple of 30. This means p-7 has to be a multiple of 30 too.

So try this algorithm:
- Take a multiple of 30. Call this "x".
- Check to see if x+7 is prime
- If it is, check to see if x+37 is too
- If they both are prime, see if they are
consecutive primes, and if there's another
prime ending in 7 "near by" too
- If not, go on to the next multiple of 30.

Starting with 30...
- 37 is prime, and so is 67. But we know there are primes in between.
- Ditto with 67 and 97
- Ditto with 97 and 127
- Ditto with 127 and 157
- 157 is prime, but 187 isn't
Keep continuing on like this

This still requires some "brute force" tests but at least it cuts down on some time.

Answer 3

It's the group 1627, 1637, 1657.

I am not sure there is an elegant way to calculate this. it may just have to be a computer program that goes through the prime numbers sequentially and tests the last digit.