Question 1207394
<pre>

Let M = the number of markers in each a box
Let C = the number of crayons in each a box
Let N = the number of boxes.

So N(M+C) must equal the total number of markers and crayons.
N(M+C) = NM + NC = 45 + 60, so N must be a factor of both 45 and 60 

For N to be as large as possible, it must be the greatest common factor
of 45 and 60, which is 15.  

N(M+C) = 15M + 15C and 15M=45 and 15C=60, or M=3 and C=4 

So there are 15 boxes, with 3 markers and 4 crayons in each box.

Edwin</pre>