Re: Australian Maths Competition 2013
	
		
	
	
		
		
			Aren't most of the questions involving integers just trial and error? The only way that I attempt them is to substitute in numbers until I get the answer, although this takes a very long time.
		
		
	 
None of the questions are ever intended to be solved by trial and error, though a bit of experimenting with numbers can often set you on the right track  
 
For the 37 & 18 question, note first that 37 x 3 = 111 which seems to be heading vaguely in the right direction.
Then 999 is also a multiple of 37, so as well as 100 being 11 less than a multiple of 37, 1000 is 1 more than a multiple of 37.
Then looking just at the remainders after division by 37:
1 leaves 1 ;  10 leaves 10 ;  100 leaves 26 (effectively -11) ;  1000 leaves 1 again, so the pattern repeats:
10,000 leaves 10 ;  100,000 leaves 26 or -11 ;  1,000,000 leaves 1 remainder
Then you just need to combine as few of these remainders as possible to get an overall remainder of 18