How to solve greater & less / inequal equation

Question

Two farmers, Eric and Josef where talking. "How many sheeps do you have?" asked Eric. "If I divide my sheeps in $2, 3, 4, 5,$ or $6$ groups, there will always be 1 sheep left." Josef answered. How many sheeps did Josef have if he had more than $100$, but less than $150$?

How do you solve this with an equation then?

The answer is $121$.

Answer 1

all equations are the same and they say $x-1=x-1$ after cancelling the common numerators, therefore we get all solutions $x$ with $100<x<150$

Answer 2

We have$$x\equiv1\pmod2$$ $$x\equiv1\pmod3$$ $$x\equiv1\pmod4$$ $$x\equiv1\pmod5$$ $$x\equiv1\pmod6$$

Using Chinese remainder theorem we have $$x\equiv1\pmod{60}\implies x=60k+1$$ where $k\in\mathbb N$

Now I'll leave it to you!

Answer 3

Because the remainder is always $1$, we need to find $ n+1,$ where n divides $2\cdots6$. It is a factor of the LCM, that is, $60$. Becuase $2\cdot60+1$ is the only number in the range $100\le x \le 150$, $121$ is your answer.