q = α–2 and pq = –α – 1

(p + q)2 = p2 + q2 + 2pq,

Thus (α–2)2 = p2 + q2 + 2(–α– 1)

p2 + q2  =  α2 – 4α + 4 + 2α + 2

p2 + q2  =  α2 – 2α + 6

p2 + q2 = α2 – 2α + 1 + 5

p2 + q2 = (α – 1)2 + 5

Thus, minimum value of p2 + q2 is 5.

We get 3x + 7y + z = 120 and 4x + 10y + z = 164.50. Subtracting, we get x + 3y = 44.50 or 2x + 6y = 89. Substitute in first equation to get x + y + z = 120 – 89 = 31.

X ->    a = 300, d = 30, t = 10;                   

    s = 5(600 + 9 × 30 × 12) = 870 × 5 × 12 = 52,200.       

    Y ->    a = 200, d = 15, t = 20; s = 10(400 + 19 × 15) × 6 = 41,100.

(a + b + c + d)2 = (4m + 1)2

Thus, a2 + b2 + c2 + d2 + 2(ab + ac + ad + bc + bd + cd) = 16m2 + 8m + 1

a2 + b2 + c2 + d2 will have the minimum value if (ab + ac + ad + bc + bd + cd) is the     maximum.

This is possible if a = b = c = d = (m + 0.25) …….since a + b + c + d = 4m + 1

In that case 2(ab + ac + ad + bc + bd + cd) = 12(m + 0.25)2 = 12m2 + 6m + 0.75      Thus, the minimum value of a2 + b2 + c2 + d2 = (16m2 + 8m + 1) – 2(ab + ac + ad +           bc + bd + cd)

= (16m2 + 8m + 1) – (12m2 + 6m + 0.75)

= 4m2 + 2m + 0.25

Since it is an integer, the actual minimum value = 4m2 + 2m + 1

If y = 2 (it cannot be 0 or 1), then x can take 1 value and z can take 2 values. Thus with y = 2, a total of 1 X 2 = 2 numbers can be formed. With y = 3, 2 X 3 = 6 numbers can be formed. Similarly checking for all values of y from 2 to 9 and adding up we get the answer as 240.