Starting from 37, adding 5 before each previous term, forms the following sequence: \[37,537,5537,55537,555537,...\] How many prime numbers are there in this sequence?
Problem
Source: 2013 Taiwan TST
Tags: number theory, prime, prime numbers, Taiwan, Taiwan TST 2013
12.07.2015 16:06
Just the first one. If $a_n$ denotes the $n$th term of the sequence, then $3 \mid a_{3k+2}$, $7 \mid a_{6k+3}$, $13 \mid a_{6k+4}$, $37 \mid a_{3k+1}$.
15.04.2021 12:37
Sorry for bumping a 5 year old thread, but was there any motivation behind checking for $7,13,37$? Or was it just a matter of trying all primes until something worked?
15.04.2021 18:51
The number of $5`s $ is nonzero multiple of $3. $ It ain't a prime. Apologies, th I wrote. Also, $a_{6k+4} $ is a subset of $a_{3k+1}. $ But there remains are the elements of $x_{6k} . $ There will be infinitely many primes, $13\nmid 55555537.$ $37$ ain't a multiple of $13, \, 555555$ is.
23.05.2021 09:19
L567 wrote: Sorry for bumping a 5 year old thread, but was there any motivation behind checking for $7,13,37$? Or was it just a matter of trying all primes until something worked? I think the simplest way is to simply write out the first few (well, you might need to write out five or six of them to see the pattern) and factorize all of them.