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Question: Suppose

*a, a+d, a+2d, ..., a+(n-1)d*are primes. Find the largest possible

*n*, under the condition that

*d*cannot be a mulitple of 10.

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 877 881 883 887 907 911 919 929 937 941 947 953 967 971 977 983 991 997

(Table of primes up to 1000)Answer: Lets make a guess. We claim the largest possible

*n≥5*.

*a*must be odd, otherwise

*a+2d*is an even number greater than 2 and thus not a prime.

If

*d*is odd, then

*a+d*is an even prime greater than 2 and thus not a prime.

If

*d*is even and not a multiple of 10, then one of

*a, a+d, a+2d, a+3d, a+4d*have the last digit being 5 and hence it must be 5. Note that there is only one odd prime, i.e. 3, which is smaller than 5. Therefore it is only possible if

*a=5*or

*a+d=5*.

If

*a+d=5*then

*a=3, d=2*. The sequence is just 3,5,7.

The only case left is that

*a=5*and then

*a+5d*is a multiple of 5. Therefore

*n*is at most 5.

You can try to search using the table.

5, 11, 17, 23, 29 is a prime sequence with

*d=6*.

5, 17, 29, 41, 53 is a prime sequence with

*d=12*.

Therefore the largest possible

*n=5*.

What if

*d*is allowed to be a multiple of 10?

No matter how large

*n*is, we can construct a sequence

*a, a+d, a+2d, ..., a+(n-1)d*of primes. It is the famous Green-Tao theorem, proved by Ben Green and Terence Tao in 2004.