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This number 25^4 has a long sequence of positive consecutive integers, the first and the last terms of which are both prime numbers. The difference between them being a divisor of 25^4. What are the first term, last term and the number of terms of this sequence? Any help would be great. Thank you.

 Sep 19, 2019
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Since the common difference of the sequence is 1, therefore this difference is = number of terms. Will try the common formula for arithmetic sequence, namely:
N / 2 * [2*F + D*(N - 1) ] for various divisors of 25^4, which are:1  5  25  125  625  3125  15625  78125  390625  >>Total = 9
From the above divisors, we find only one of them, 625, meets the conditions in the question:
625 / 2 * [2*F + 1*(625 - 1) ], solve for F.
F = 313 - This is the first term and is also a prime.
313 + 625 - 1 = 937 - This is the last term and is also a prime.
[313 + 937] / 2 * 625 =25^4 - which checks out.

 Sep 20, 2019

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