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 250x+750y=3000 ,where x and y are positive integers.Dividing each side of 250x+750y=3000 by 250 gives x+3y=12, where x and y are postive integers.Since 3 and 12 are each divisible by 3,it follows that x=12−3y must also be divisible by 3.If x=3 then y=3;ifx=6,then y=2 ;and if x=9 then y=1.If x=12,then 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