Find the least value of x which when divided by 3 leaves remainder 1, ... A number when divided by 3 gives a remainder of 1; when divided by 4, gives a remainder of 2; when divided by 5, gives a remainder of 3; and when divided by 6, gives a remainder of 4. Find the smallest such number.
How to solve this question in 1 min?
Any help would be appreciated. :)

So we need $\displaystyle x=3a+1=4b+2=5c+3=6d+4$
which can also the written as $\displaystyle x=3(a+1)-2=4(b+1)-2=5(c+1)-2=6(d+1)-2$
So, we need to find $x$ such the remainder $=-2$ for the divisors $3,4,5,6$
Now, the smallest number which is divisible by $3,4,5,6$ is lcm$(3,4,5,6)=60$
So, $60m-2$ (where $m$ is an integer) will leave $-2$ as remainder
Find proper $m$ for the minimum positive value of $x$
