Arithmetic Progression With Primes

Relevant wiki: Arithmetic and Geometric Progressions Problem Solving

Before starting: Assume the Arithmetic Progression (AP) has its first term a and its difference d. So its general term will be:

$t_{n} = a + (n - 1)d.$

Since all primes (except for 2) are odd, therefore the difference d between any 2 successive terms should be even or a multiple of 2.

Since any consecutive 3 terms must be prime, none of them should be divisible by 3. Therefore, the difference d between any 2 successive terms should be a multiple of 3.

Since any consecutive 5 terms must be prime, none of them should be divisible by 5. Therefore, the difference d between any 2 successive terms should be a multiple of 5.

Since any consecutive 7 terms must be prime, none of them should be divisible by 7. Therefore, the difference d between any 2 successive terms should be a multiple of 7.

Therefore d should be a multiple of 2 X 3 X 5 X 7 = 210.

Therefore d should be in the form 210k where k is a positive integer.

Since all terms are less than 3000, therefore:

$t_{10} = a + 9 d < 3000$

a + 9 X 210 k < 3000 a + 1890 k < 3000 1890 k < 3000 k < 1.587 k = 1 That means d = 210.

Since a + 1890 < 3000 a < 3000 – 1890 a < 1110.

Conclusion: This AP has the first term is a prime number less than 1110 and the difference is 210.

The general term of this AP is : a + (n – 1)d = a + (n – 1) X 210.

$t_{n} = a + 210 n - 210.$

Since all terms are prime so none of the numbers should be a multiple of 11 except if the first term a = 11. Therefore :

$t_{n}$ mod 11 > 0 for n > 1.

For n = 1, we can have a = 11. In this case,

$t_{2} = 11 + 210 = 221 = (13 X 17)$ which is composite and not prime. This should be rejected.

We have now:

(a + 210 n – 210) mod 11 > 0 for n = 1, 2, 3, … 10.

(a + 210(n – 1)) mod 11 > 0 for n = 1, 2, 3, … 10.

(a + n – 1) mod 11 > 0 for n = 1, 2, 3, … 10.

That means a, a + 1, a + 2, … a+9 are not multiples of 11. That happens only if a + 10 is a multiple of 11 or a mod 11 = 1. That means a = 11 k + 1.

Since a < 1110 11 k + 1 < 1110 11 k < 1109 k < 100.8, therefore k = 1, 2, 3, … , 100.

Also note that when k is odd a is even and vice vera. Since we want a to be odd, k has to be even. We can rewrite the formula for a in the form: a = 22 m + 1 where m = 1, 2, 3, … 50

In that case:

$t_{n} = 22 m + 1 + 210 n - 210 = 22 m + 210 n - 209.$

Since /(t_{n}/) mod 13 > 0, therefore (22 m + 210 n – 209) mod 13 > 0.
(9m + 2n – 1) mod 13 > 0 for m = 1,2,3 … 50 and for n = 1,2,3 … 10

Substituting n by the values 1,2,3 to 10 we get: 9m mod 13 > 0, 9m + 1 mod 13 > 0, 9m + 2 mod 13 > 0, 9m + 3 mod 13 > 0, 9m + 4 mod 13 > 0, 9m + 5 mod 13 > 0, 9m + 6 mod 13 > 0, 9m + 7 mod 13 > 0, 9m + 9 mod 13 > 0, and 9m + 11 mod 13 > 0. (for m = 1,2,3 … 50)

That means the only possibles values for m will be when : 9m + 8 mod 13 = 0, 9m + 10 mod 13 = 0, and 9m + 12 mod 13 = 0. (for m = 1,2,3 … 50)

This will give us the possible values for m: Since 9m + 8 mod 13 = 0 gives us m = 2, 15, 28 and 43. 9m + 10 mod 13 = 0 gives us m = 9, 22, 35 and 48. 9m + 12 mod 13 = 0 gives us m = 3, 16, 29 and 44.

Since a = 22m + 1, these 12 values of m will give us the following 12 values for a (in order): a = 45, 67, 199, 331, 353, 485, 617, 639, 771, 947, 969, 1057

Excluding the composite numbers we get the possible values for a to be prime are: a = 67, 199, 331, 353, 617, 947.

If a = 67: AP will be: 67, 277, 487, 697 (= 17 X 41) to be rejected

If a = 331: AP will be: 331, 541, 761, 971, 1181, 1391 (= 13 X 107) to be rejected

If a = 353: AP will be: 353, 563, 773, 983, 1193, 1403 (= 23 X 61) to be rejected

If a = 617: AP will be: 617, 827, 1037 (= 17 X 61) to be rejected

If a = 947: AP will be: 947, 1157 (= 13 X 89) to be rejected

The only one which works is when a = 199: AP will be: 199, 409, 619, 829, 1039, 1249, 1459, 1669, 1879, 2089

In this case: the total will be: $\frac{10}{2}$ X (199 + 2089) = 11440

Answer = $\boxed{11440}$