This is a short number puzzle:
Find a number $x$, which leaves remainder $1$ when divided by $2,3,4,5$ and $6$.
Here is the some more explanation:
$x$ divided by $2$ gives remainder $1$
$x$ divided by $3$ gives remainder $1$
$x$ divided by $4$ gives remainder $1$
$x$ divided by $5$ gives remainder $1$
$x$ divided by $6$ gives remainder $1$
Brute force, everyone.
There are 1.667 correct answers just between 1 and 100.000
You may check this fiddle:
https://jsfiddle.net/g4tk05pn/
So while the question is not spesific like smallest number possible, there is not just one answer.
The answer is simple, it's 61.
LCM of 2,3,4,5,6 (which is 60), plus 1.
The Chinese remainder theorem renders this type of problem easy. However, in this case the answer is even easier:
The answer is $1$, since:
$$ 1~\text{mod}~2 \equiv 1 \\ 1~\text{mod}~3 \equiv 1 \\ 1~\text{mod}~4 \equiv 1 \\ 1~\text{mod}~5 \equiv 1 \\ 1~\text{mod}~6 \equiv 1 $$
The Chinese remainder theorem tells us that there are an infinite number of numbers satisfying this type of system of equations. There is a bit of difficulty in that our moduli are not pairwise coprime, so we must take the least common multiple of the moduli instead of just multiplying them:
$\text{LCM}(2,3,4,5,6)=60$, so our final answer is $x\equiv 1~\text{mod}~60$
It's 721, or in other words, 6! + 1.
