Arithmetic Progression of Primes of length 7

Question

Determine the least possible value of the largest term in an arithmetic progression of seven distinct primes.

I don't have a great understanding of how to start the problem. I have seen approaches of searching for sequences with a difference between terms of $(2\cdot3\cdot5\cdot7)$ and then seeing what the largest number is.

How would you go about starting this question? Is it possible to bound the largest term?

Edit - commentators mention that we can search for multiples that differ by 30 instead of 210. Why is this the case?

Answer 1

I had found an AP which has all prime numbers for now: 7, 157, 307, 457, 607, 757, 907. (This IS the actual answer too, which future me will say in a few lines.)

The common difference is 150. So, the answer is lesser than or equal to 907. If you wanted a more procedural method, then maybe you could find things like:

1. The AP cannot start with any number lesser than 7 (say k), since the (k+1)th term (k + kd) = k(d + 1) is definitely divisible by k.
2. The AP has to start with a prime greater than or equal to 7 (an implication of 1.)
3. The common difference cannot be odd (It should be a 2 multiple).
4. Any prime (greater than or equal to 7) mod 3 gives 1 or 2. So, if the common difference is not a 3 multiple, then at least one term would be a 3 multiple. Example: If the starting prime is of the form 3k + 1, and the common difference is of the form 3m + 2, then the second term would be a 3 multiple.

You can also generalize the fourth statement to other numbers.

Edit: Wait, it IS 907! You can prove this easily. Remember how I said you can generalize the fourth statement? Well, you can do that with 5 too. (That is, the common difference is a 5 multiple.) So, the common difference HAS to be a 2 * 3 * 5 = 30 multiple.

After trying with 7 from common difference 30 to 150, you can see that common difference 150 works (With final number as 907). Now, we have to try with other primes.

But, when we try with 11, we always get a 7 multiple in between. Why is that? Well, 7 is of the form 7m, and 30 is of the form 7k + 2. So, only the 8th term (7m + (7k + 2) * 7) can be a 7 multiple. But, 11 is of the form 7m + 4, and 30 is of the form 7k + 2. So, the 5th term would be a 7 multiple. (7m + 4 + 35k + 10 = 7m + 35k + 14). The same happens to all 30 multiples from 30 to 180. It doesn't have a 7 multiple only when the common difference is a 7 multiple (210 is the least value for common difference now). The least possible end term is 11 + 210 * 7 = 1481, way higher than 907. And, the following primes are not 7 multiples either, so they also can have minimum common difference as 210, which still has a higher end term. So, we can see that 907 is the answer!

(Note: Please notify me if I am wrong, since I can at least see what mistake I did in my calculations.)