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This number 15^6 has a long sequence of positive consecutive integers. The last term is a 4-digit prime number whose first digit(from the left) x second digit x third digit - fourth digit = 101. What are the first term, last term and the number of terms of this sequence? Thank you for help.

 Aug 22, 2019
 #1
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Pls check your problem for typos, because 156 = 11,390,625. 

 

Not only is 0625 not prime, but all the other conditions fail too.

 

Maybe I'm misunderstanding the problem?  Unlikely.

.

 Aug 22, 2019
 #2
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The sequence doesn't start at 1 and end at 15^6. It starts somewhere else and ends in a 4-digit integer, which is a prime number. You didn't understand the question.

 Aug 22, 2019
 #4
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That's right, I didn't understand the question.  But I understand English just fine.  The question is inadequately stated.

Guest Aug 23, 2019
 #3
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Since there are over 1000 prime numbers between 1,000 and 10,000, I know of no formula to narrow this down to one or two numbers except by a short computer code that will eliminate 99% of them. I wrote a short code that narrowed them down to just 3: 6,367,  6,637,  9,437. By a simple test using this modified AP formula, it turn out that the last term is: 6,637.
  N/2 * ((4*15^6 / N) - (2*6637) + (1*(N-1)) = 15^6, Solve for N
Plugging it in the above formula and using the quadratic formula, it gives N=2,025 which is the number of terms;
6,637 - 2,025 + 1 =4,613 which is the first term.
[4,613 + 6,637] / 2 * 2,025 =15^6 which checks out.

 Aug 22, 2019

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