PK 2015 Q19
19. How many 2-digit numbers can be written as the sum of exactly six different powers of 2, including \(2^0\) ?
- A 0
- B 1
- C 2
- D 3
- E 4
Pink Kangaroo 2015, Question 19. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 2
Since \(2^7 = 128\) is larger than 100, the only powers we can choose from are the first seven powers: \(2^0\) to \(2^6\), i.e. 1, 2, 4, 8, 16, 32, 64. The sum of all seven is 127. We need to eliminate one of these, and be left with a total under 100, so the only possibilities to remove would be 32 or 64. Hence there are two options:
1 + 2 + 4 + 8 + 16 + 32 = 63, and 1 + 2 + 4 + 8 + 16 + 64 = 95.
Solution © UK Mathematics Trust, from the PK 2015 solutions.