David has an interesting habit of spending money. He spends exactly £X on the Xth day of a month. For example, he spends exactly £5 on the 5th of any month. On a few days in a year, David noticed that his cumulative spending during the last 'four consecutive days' can be expressed as $$2^N$$ where N is a natural number. What can be the possible value(s) of N?

The sum of 4 consecutive numbers should be a power of 2. 
The powers of 2 are 1, 2, 4, 8, 16, 32, 64, 128. The maximum possible value that 4 consecutive days can take is 28 + 29 + 30 + 31 = 118.
We can eliminate 1, 2, 4, and 8 since sum of 4 consecutive integers is always greater than 9. 
Let the first day be x. 
x+x+1+x+2+x+3 = 16
=> 4x = 10
x = 2.5
We can eliminate 16.
Let us check for 32.
x+x+1+x+2+x+3 = 32
4x = 26.
x is not an integer. 
Let us check the case in which x is the last day of the month. 
Even in the month of February, the least value that x can take is 28.
28+1+2+3 = 34 > 32.
We can eliminate 32 as well. 
Let us check for 64.
4x+6 = 64
4x = 58
The sum of no 4 consecutive days in the same month can be expressed as 4k+6. 
Let us check the cases in which 2 months are involved.
29+30+1+2 = 62 < 64.
30 + 31 + 1 + 2 = 64. This is a possible combination. 
There are 7 months with 31 days in a year. We have to eliminate December since 1 will spill over to the 1st of January of the next year. 
Therefore, in a year, there will be 6 such instances. Therefore, option B is the right answer.