a) The highest common factor of 75, 105 and 165 is 15. Therefore, the sum of Tommy‘s purchases must be a multiple of 15. The largest multiple of 15 that is less than 700 is 690. When 690 is divided by 15, the quotient is 46. The respective quotients when 75, 105 & 165 are divided by 15 are 5, 7 & 11. Since it is possible to find a combination of multiples of 5, 7 & 11 which add to 46 (namely 2x(5+7+11)=46), then it is possible to purchase items which total $6.90 (namely (2x $0.75)+(2x $1.05)+(2x $1.65)).

In equation/inequality form:

75a + 105b + 165c = 15m < 700

5a + 7b + 11c = m < 46.7 Therefore m = 46

By observation, 2x(5+7+11) = 46

Therefore, (2x $0.75)+(2x $1.05)+(2x $1.65) = $6.90

b) To maximise the number of incorrect responses which scored zero, the total of 35 should comprise the maximum number of correct answers worth 8 points. As it happens, when 35 is divided by 8 the quotient & remainder are 4 r 3, so Vicki could score 35 with four correct answers, one no response and seven incorrect responses. Therefore, the maximum number of incorrect responses with a score of 35 is seven.

c) The highest common factor of 85 and 51 is 17. Steve can produce any multiple of 17 by repeatedly pressing +, +, ÷ (+51+51-85 = +17). Since he can only ever add or subtract a multiple of 17, all of the numbers he is able to produce are multiples of 17. When 2003 is divided by 17, the quotient and remainder are 117 r 14. Since 14 is closer to 17 than to zero, the multiple of 17 closest to 2003 is 118 x 17 = 2006.

d) Since 4y and 96 are both divisible by 4, 3x must also be divisible by 4. Since 3 and 4 have no common factors (they are co-prime), ‘x’ must be divisible by 4. For x as each multiple of 4, including zero, the corresponding value of y is as follows:

(0,24), (4,21), (8,18), (12,15), (16,12), (20,9), (24,6), (28,3), (32,0). There are nine possible pairs of (x,y).

Perhaps someone who is currently more familiar with the appropriate notation could post the corresponding equations/inequalities for b), c) & d).