Smallest number???

The LCM of 3,4,5 and 6 is 60.

we need remainder 1 when divided by 3. 3-1=2.

Similarly, 4-2 = 5-3 = 6-4 = 2.

So if we subtract 2 from 60, we get 58 which satisfies all 4 conditions.

The 4. condition is completely redundant. 1,2 already imply it.

$n\equiv 1\pmod{\!3}\,\Rightarrow\, n=3k+1$ .

$n\equiv 3k+1\equiv 2\pmod{\! 4}\iff -k\equiv 1\pmod{4}$

\stackrel{:(-1)}\iff k\equiv -1\equiv 3\pmod{\!4}\,\Rightarrow\, k=4h+3

$n=3(4h+3)+1=12h+10$ .

$n\equiv 12h+10\equiv 2h\equiv 3\equiv -2\pmod{\!5}$

\stackrel{:2}\iff h\equiv -1\equiv 4\pmod{\! 5}\,\Rightarrow\, h=5m+4

$n=12(5m+4)+10=60m+\boxed{58}$ .

We never used 4. and we still have $n\equiv 60m+58\equiv 4\pmod{\!6}$ .

Please ask if I expressed something unclearly :)

get the LCM of 3,4,5, and 6 = 60,subtract the common difference of the remainder = 2 60 - 2 = 58