1. 2

2. 6

3. Option: 1

4. Option: 1

**Explanation**:

a. Minimum possible number is when there is 1 of type a and 99 of type b which is in accordance to the condition.

b. Maximum possible number is when 1 of a, 2 of b, 4 of c, 8 of d, 16 of e, 32 of f,

Now left boxes woud be 100 – (1+2+…32) = 37

Now if one more type is to be added then we need at least 64 which is not available thus maximum possible is 6.

c. Lets try to prove the given options possible using easy numbers.

op1:never possible

op2: 1,30,69 is possible

op3: 1,2,4,18.75 is possible

op4: 1,9,30,60 is possible.

d. There have to be then at least 31 + `1 + 43 = 75 gifts of same type,

Thus maximum possible number of boxes = 5 when all types are lest 1,2,4,18,75