Find the least number which when divided by 2, 3, 4, 5, 6 leaves a remainder of 1 but it is divided by 7 completely.

You definitely need $n\equiv 1\pmod {60}$ and $n\equiv 0\pmod 7$. So the trick is to apply the Chinese remainder theorem. Solve $60x+7y=1$ with $(x,y)=(2,-17)$.

Then $n\equiv 1\cdot 7\cdot (-17)+0\cdot 60\cdot 2\pmod{420}$, or $n\equiv -119\pmod{420}$. The smallest such positive number is $420-119=301$.

Consider $60k+1,$ $$60(7n)+1=420n+1$$ $$60(7n+1)+1=420n+61$$ $$60(7n+2)+1=420n+121$$ $$60(7n+3)+1=420n+181$$ $$60(7n+4)+1=420n+241$$ $$60(7n+5)+1=420n+301$$ $$60(7n+6)+1=420n+361.$$ Now by divisibility test for 7, among $1, 61, 121, 181, 241, 301, 361$, only $301$ is divisible by $7.$

Therefore any number of the form $420n+301$ satisfies the given requirements.

Here's another way to tackle it. You already figured out that you're looking for $60k+1$. When you multiply $60$ by $k$ you want a predecessor to a multiplication of $7$.

$60 \equiv 4 \pmod 7$ and $5 \times 4 = 20$ which is a predecessor to a multiplication of $7$.

So $k=5$.