SAT Mathematics Practice Test 2

Let x be the number of 250 bonuses awarded , and let y be the number of 750 bonuses awarded.Since 3000 in bonuses were awarded,and this included at least one 250 bonus and one 750 bonus,it follows that $250\mathit{x}+750\mathit{y}=3000$ ,where $\mathit{x}$ and $\mathit{y}$ are positive integers.Dividing each side of $250\mathit{x}+750\mathit{y}=3000$ by $250$ gives $\mathit{x}+3\mathit{y}=12,$ where $\mathit{x}$ and $\mathit{y}$ are postive integers.Since $3$ and $12$ are each divisible by 3,it follows that $\mathit{x}=12-3\mathit{y}$ must also be divisible by 3.If $\mathit{x}=3$ then $\mathit{y}=3$;if$\mathit{x}=6$,then $\mathit{y}=2$ ;and if $\mathit{x}=9$ then $\mathit{y}=1$.If $\mathit{x}=12$,then $\mathit{y}=0$,but this is not possible since there was at least one 750 bonus awarded.Therefore,the possible numbers of 250 bonuses awarded are 3, 6 and 9.Any of the numbers 3,6 and 9 may be gridded as the correct answers