# Consecutive number division puzzle [closed]

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 $$2,3,5,7$$.