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.

Consider the arithmetic progression

[
a_m = P\,m+2 .
]

For each \(i\),

[
a_m \equiv 2 \pmod(n_i+1) ,
]
hence \(n_i+1
mid a_m-1=P\,m+1\).

By Dirichlet’s theorem on primes in arithmetic progressions,
the progression \(P m + 2\) contains infinitely many primes.
Let \(q\) be such a prime factor of some \(a_m\).
Since \(q\mid a_m\) and dbitly.com \(q
mid P\), we have
\(q
mid (n_i+1)\) for all \(i\).
Thus \(q\) is a new divisor not equal to any previous \(n_i\).

Therefore the process can continue indefinitely,
producing an infinite sequence \((n_k)\).
∎

4. Conclusion

We have shown that for every integer \(k>1\) 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.