Dianabol Only Cycle Dianabol Only Cycle

Answer – General statement



Let \(k\geq 1\) be an integer and let




[
n_1,\,n_2,\,\dots ,\,n_k\in \mathbb Z
]



be any \(k\) integers.

For every real number \(X>0\) there exists a prime \(p>X\) such that



[
p\mid (\,n_i+1\,)\quad\textfor some i\;(1\le i\le k).
]



Equivalently, the set




[
\bigcup_i=1^k\,\textprimes dividing (n_i+1)\,
]



is infinite.



---




Proof


Let \(P=\prod_i=1^k(n_i+1)\).

Choose an integer \(m\) larger than any prescribed bound.




C>1\) there exists a
\(k\)-digit number with all digits different.
The construction of the sequence
\(\,N_k\,\) guarantees that such numbers exist for all
values of \(k\).



Hence there are infinitely many integers whose decimal representation
contains only distinct digits.

These are precisely the numbers \(1,2,\dots ,9,10,12,\dots ,98,102,
103,\dots \), and so on.



∎



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5. Remarks




The set of such numbers is not a perfect arithmetic progression
(there are gaps, e.g. between \(99\) and \(100\)).
In base‑\(b\) there can be at most \(b^\,b\) distinct‑digit numbers.
These integers are sometimes called pandigital* in the sense of
using each digit only once, but not necessarily all digits.



The proof above shows that, for any finite alphabet, one can construct
infinitely many words with no repeated symbols.