An competitive question asked this time

[varun atri]

a boy running up a stair case finds when he goes up two steps at a time there is one step over when he goes up three at a time there are two over and when he goes up four at a time, there are three over. find number of stairs which is b/w 40 & 50.

a.) 47 b.)45 c.)42 d.)49 e.) none
:rofl:

 

[HallsofIvy]

Why is this "competative"? It is a pretty straightforward "Chinese remainder theorem" problem. It is the same as asking for x such that:
x= 1 (mod 2), x= 2 (mod 3), and x= 3 (mod 4).

If x= 3 mod 4, then x= 4i+ 3 for some positive integer i. Putting that into the second equation, x= 4i+ 3= 2 (mod 3) so that 4i= -1= 2 (mod 3). i= 2 (mod 3). Continue.