LCM !!

If the number leaves a remainder of $1$ when divided by $2,3,4,5\;and\;6$ then it leaves a remainder of 1 when divided by $LCM(2,3,4,5,6) = \boxed{60}$ .So the number must be of the form $60x+1$ and is divisible by $7$ .The least number satisfying these conditions is $60(5)+1=300+1=\boxed{301}=7\times43$

Well we can also do this by heat and trial method too, just have to look at some points: 1)The no. leaves no reminder when divided by seven so it is a multiple of 7.

2)It would be odd. since it leaves remainder of 1 when divided by 2

3)Since it leaves remainder of 1 when divided by 5 then the unit place of the no. is also 1.

4)by "3)"we can also conclude that the no.=n 3 * 7 = x 1 where n is any no. with unit place 3 and n is the resultant of product with unit place 1.

5)The n 3 would be prime.

By keeping all these 5 points in mind . I got the answer = 43*7=301