This number 65^4 has a long sequence of positive consecutive 4-digit integers. The first term, the 3001st term and the last term are all 4-digit prime numbers. What is the first term, the last term and the number of terms of this sequence? Thank you for help.

Guest Aug 13, 2019

#1**0 **

**There are 1,061 primes between 1,000 and 9999. 289 of them have a difference of 3,000. Only one of them meets the conditions of the sequence. By the process of elimination, through a computer code, the first term of the sequence =2,113. The 3001st term =2113 + 3000 =5,113. Both of these numbers are primes. 65^4 = N/2 * [2 * 2113 + (N - 1)*1, solve for N Using quadratic formula, we have: N = 4,225 - This is the number of terms. 2,113 + 4,225 - 1 =6,337 - This is the last term [also a prime] [2,113 + 6,337] / 2 * 4225 = 65^4 - This checks out. **

Guest Aug 14, 2019

#2**+1 **

**This number \(65^4\) has a long sequence of positive consecutive 4-digit integers. The first term, the 3001st term and the last term are all 4-digit prime numbers. What is the first term, the last term and the number of terms of this sequence?**

\(\text{$a_1$ is a prime number hence odd } \\ \text{$a_n$ is a prime number hence odd } \)

\(\begin{array}{|rcll|} \hline && a_n = a_1+(n-1) \\ &or& n = a_n-a_1+1 \qquad \text{$a_n$ is odd and $a_1$ is odd hence $n$ is odd} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 65^4 &=& a_1 + (a_1+1)+ (a_1+2) + \ldots + \Big(a_1+(n-1) \Big) \\ 65^4 &=& na_1 + \dfrac{ \Big(1+(n-1)\Big)}{2} (n-1 ) \\ 65^4 &=& na_1 + \dfrac{n(n-1)}{2} \\ 65^4 &=& n\left( a_1 + \dfrac{n-1}{2}\right) \\ 65^4 &=& n\left( \dfrac{2a_1+n-1}{2}\right) \quad | \quad n = a_n-a_1+1 \\ 65^4 &=& (a_n-a_1+1)\left( \dfrac{2a_1+a_n-a_1+1-1}{2}\right) \\ \mathbf{65^4} &=& \mathbf{(a_n-a_1+1)\left( \dfrac{ a_1+a_n }{2}\right)} \quad | \quad \text{$a_1+a_n$ is even! $\quad a_n-a_1+1=n$ is odd!} \\ \mathbf{65^4} &=& \mathbf{x\times y} \\ && \boxed{x = a_n-a_1+1},\ \boxed{y = \left( \dfrac{ a_1+a_n }{2}\right)} \quad \text{$x$ and $y$ are divisors of $65^4$ } \\\\ \hline x &=& a_n-a_1+1 \quad \text{ or } \\ \mathbf{ a_n } &=& \mathbf{x+p-1} \\\\ y &=& \left( \dfrac{ a_1+a_n }{2}\right) \\ y &=& \left( \dfrac{ a_1+x+p-1 }{2}\right)\quad \text{ or } \\ \mathbf{a_1} &=& \mathbf{y-\dfrac{x-1}{2}} \\ \hline \end{array}\)

**The Divisors of \(65^4\) are: **

\(\text{1 | 5 | 13 | 25 | 65 | 125 | 169 | 325 | 625 | 845 | 1625 | 2197 | 4225|$\\$ | 8125 | 10985 | 21125 | 28561 | 54925 | 105625 | 142805 | 274625 |$\\$ |714025 | 1373125 | 3570125 | 17850625 (25 divisors)}\)

\(\begin{array}{|rr|cc|cc|c|} \hline & & & \text{prime} &&\text{prime} \\ x & y & a_1 = y-\dfrac{x-1}{2} & \text{number} & a_n = x+a_1-1 & \text{number} & n =a_n-a_1+1 \\ \hline 1 & 17850625 & a_1 > 9999 \\ 5 & 3570125 & a_1 > 9999 \\ 13 &1373125 & a_1 > 9999 \\ 25 & 714025 & a_1 > 9999 \\ 65 & 274625 & a_1 > 9999 \\ 125 & 142805 & a_1 > 9999 \\ 169 & 105625 & a_1 > 9999 \\ 325 & 54925 & a_1 > 9999 \\ 625 & 28561 & a_1 > 9999 \\ 845 & 21125 & a_1 > 9999 \\ 1625 & 21125 & a_1 > 9999 \\ 2197 & 10985 & a_1=7027 & \text{yes} & a_n= 9223 & \text{no} \\ 4225 &4225& \color{red}a_1 = 2113 & \text{yes} & \color{red}a_n= 6337 & \text{yes} & \color{red}n= 4225 \\ \hline \end{array}\)

Here is only one solution so \(a_{3001} \)must be a prime number. \((a_{3001}=a_1+3000 = 2113+3000=5113\text{ prime number})\)

\(a_{3001} \) is irrelevant for the calculation!

heureka Aug 14, 2019