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This suggested edit was approved and applied to the post over 1 year ago by WheatWizard‭.

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Find near miss prime multiples.
  • Given a number $n \geq 3$ as input output the smallest number $k$ of modular residues of $k$ by the first $n$ primes is exactly $\{-1,0,1\}$.
  • That is there is a prime in the first $n$ primes that divides $k$, one that divides $k+1$ and one that divides $k-1$, *and* every prime in the first $n$ primes divides one of those three values.
  • For example if $n=5$, the first $3$ primes are $2, 3, 5, 7, 11$ the value must be at least $10$ since with anything smaller $11$ is an issue. So we start with $10$:
  • $$
  • 10 \equiv 0 \mod 2
  • $$
  • $$
  • 10 \equiv 1 \mod 3
  • $$
  • $$
  • 10 \equiv 0 \mod 5
  • $$
  • $$
  • 10 \equiv 3 \mod 7
  • $$
  • Since $10 \mod 7 \equiv 3$, $10$ is not a solution so we try the next number.
  • * $11 \equiv 4\mod 7$ so we try the next number.
  • * $12 \equiv 2\mod 5$, so we try the next number.
  • * $13 \equiv 3\mod 5$, so we try the next number.
  • * $14 \equiv 3\mod 11$, so we try the next number.
  • * $15 \equiv 4\mod 11$, so we try the next number.
  • * $16 \equiv 2\mod 7$, so we try the next number.
  • * $17 \equiv 2\mod 5$, so we try the next number.
  • * $18 \equiv 3\mod 5$, so we try the next number.
  • * $19 \equiv 5\mod 7$, so we try the next number.
  • * $20 \equiv 9\mod 11$, so we try the next number.
  • $$
  • 21 \equiv 1 \mod 2
  • $$
  • $$
  • 21 \equiv 0 \mod 3
  • $$
  • $$
  • 21 \equiv 1 \mod 5
  • $$
  • $$
  • 21 \equiv 0 \mod 7
  • $$
  • $$
  • 21 \equiv -1 \mod 11
  • $$
  • 21 satisfies the property so its the answer.
  • ## Task
  • Take as input $n \geq 3$ and give as output the smallest number satisfying.
  • This is code-golf. The goal is to minimize the size of your source code as measured in bytes
  • ## Test cases
  • ```
  • 3 -> 4
  • 4 -> 6
  • 5 -> 21
  • 6 -> 155
  • 7 -> 441
  • ```
  • Given a number $n \geq 3$ as input output the smallest number $k$ of modular residues of $k$ by the first $n$ primes is exactly $\{-1,0,1\}$.
  • That is there is a prime in the first $n$ primes that divides $k$, one that divides $k+1$ and one that divides $k-1$, *and* every prime in the first $n$ primes divides one of those three values.
  • For example if $n=5$, the first $5$ primes are $2, 3, 5, 7, 11$ the value must be at least $10$ since with anything smaller $11$ is an issue. So we start with $10$:
  • $$
  • 10 \equiv 0 \mod 2
  • $$
  • $$
  • 10 \equiv 1 \mod 3
  • $$
  • $$
  • 10 \equiv 0 \mod 5
  • $$
  • $$
  • 10 \equiv 3 \mod 7
  • $$
  • Since $10 \mod 7 \equiv 3$, $10$ is not a solution so we try the next number.
  • * $11 \equiv 4\mod 7$ so we try the next number.
  • * $12 \equiv 2\mod 5$, so we try the next number.
  • * $13 \equiv 3\mod 5$, so we try the next number.
  • * $14 \equiv 3\mod 11$, so we try the next number.
  • * $15 \equiv 4\mod 11$, so we try the next number.
  • * $16 \equiv 2\mod 7$, so we try the next number.
  • * $17 \equiv 2\mod 5$, so we try the next number.
  • * $18 \equiv 3\mod 5$, so we try the next number.
  • * $19 \equiv 5\mod 7$, so we try the next number.
  • * $20 \equiv 9\mod 11$, so we try the next number.
  • $$
  • 21 \equiv 1 \mod 2
  • $$
  • $$
  • 21 \equiv 0 \mod 3
  • $$
  • $$
  • 21 \equiv 1 \mod 5
  • $$
  • $$
  • 21 \equiv 0 \mod 7
  • $$
  • $$
  • 21 \equiv -1 \mod 11
  • $$
  • 21 satisfies the property so its the answer.
  • ## Task
  • Take as input $n \geq 3$ and give as output the smallest number satisfying.
  • This is code-golf. The goal is to minimize the size of your source code as measured in bytes
  • ## Test cases
  • ```
  • 3 -> 4
  • 4 -> 6
  • 5 -> 21
  • 6 -> 155
  • 7 -> 441
  • ```

Suggested over 1 year ago by trichoplax‭