Determining composite numbers by using geometric progression.

Question

I have a problem which involves finding composite numbers among given set of numbers of the form $11111...1$$(n-digits)$.

Which of the following numbers is/are composite

(i) $11111...1$$(91-digits)$

(ii) $11111...1$$(81-digits)$

(iii) $11111...1$$(75-digits)$

(iv) $11111...1$$(105-digits)$

Now I know that these numbers can be simplified using Geometric series.

$$11111...1(n-digits) = 10^0+10^1+10^2+....+10^{n-1}$$ $$= \Biggl(\frac{10^n-1}{10-1}\Biggr)$$

But I can't further simplify from here so I help for this.

I have seen a similar post in MSE Primes with digits only 1 but it was tagged under number theory which I am not familier with, my question is about a specific problem from sequence and series.It would be good if I am able to solve using various sequences and series.

This has been answered here. Those numbers are rep-units (denoted with $R_n$) , which can only be prime if $n$ is prime. Only nine primes of this form are known, five of them actually proven to be prime. — Peter, Jun 16 '19 at 12:05

score 0 · Accepted Answer · answered Jun 16 '19 at 12:13

0

All those four numbers are compsit.

For the first one we hae $$91=7\times 13$$

Thus we can factor $1111111$ and the number will be the product of at least two integers.

The other three numbers are divisible by $3$ because sum of their digits are divisible by $3$

answered Jun 16 '19 at 12:13

Mohammad Riazi-Kermani

68,728

So how can we factor the first number from the pattern i.e(how can you show that the number is product of two integers from the described pattern)? – pranjal verma Jun 16 '19 at 13:05
@prankal you get $1111111(1+10^7+10^{14}+...+10^{84}$ – Mohammad Riazi-Kermani Jun 16 '19 at 14:43
Thank you very much sir for helping me. – pranjal verma Jun 17 '19 at 09:51

Determining composite numbers by using geometric progression.

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