Find 4 consecutive numbers that divide into 2,3,5,7,respectively or prove it's impossible. No computers, except for checking. If there are multiple solutions, list them all or make a formula for generating them.


We need a number $n$ such that:

  • $n$ is a multiple of $2$;
  • $n+1$ is a multiple of $3$;
  • $n+2$ is a multiple of $5$;
  • $n+3$ is a multiple of $7$.

By the

Chinese Remainder theorem,

this problem is

solvable, and there is a unique solution modulo $2\cdot3\cdot5\cdot7=210$.

You can find

one solution by trial and error, and then the rest come naturally by adding multiples of $210$.

To find it, I would

go one by one through the congruences to be solved.
First two: $2$ divides $n$ and $3$ divides $n+1$ iff $n\equiv2$ modulo $6$.
First three: $n\equiv2$ modulo $6$ and $n\equiv3$ modulo $5$ iff $n\equiv8$ modulo $30$.
All four: $n\equiv8$ modulo $30$ and $n\equiv4$ modulo $7$ iff $n\equiv158$ modulo $210$.

So the answer is

$n=158+210k$, $k\in\mathbb{Z}$. You can check that $158,159,160,161$ are respectively multiples of $2,3,5,7$.


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