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.
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.
+26367 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!
