Problem Solving

On 1st May, Mr. X arrived in a new city and was looking for a place to stay. He met a landlord who offered to rent his apartment at a reasonable price but wanted Mr. X to pay the rent on a daily basis. Mr. X had a silver bar of 31 inches and an inch of the silver bar was exactly equal to a day's rent. He agreed to pay an inch of the silver bar towards the daily rent. Mr. X wanted to make minimum number of pieces of silver bar but did not want to pay any advance rent. How many pieces did he make?
(A)5 (B)8 (C)16 (D)20 (E)31

Mr.X can do this with 5 pieces.
On May 1st, he gives the landlord 1 inch piece.
On the 2nd day, he gives the landlord a 2 inch piece, and takes the 1 inch back.
On the 3rd, gives the landlord the 2 inch piece. (now she has 2+1)= 3 inches for 3 days.
On the 4th day, he gives the landlord the 4 inch piece, but takes back the 2 and 1 inch pieces.
In the same way he can do it with 1+2+4+8+16 = 31 inches for 31 days, with 5 pieces.

Yes, as 'eagleeye' said, 5-pieces is the answer.
Beauty of math is illustrated in this problem.
If Mr X cuts the silver bar of 31 inches in to the pieces with following sizes (in inches), it will result in minimum number of pieces- solution:
1, 2, 4, 8, 16
On day-1 Mr X gives 1-inch piece to the landlord
On day-2 he gives 2-inch piece and takes back the 1-inch piece
On day-3 he gives 1-inch piece in addition to the already given 2-inch piece, thus making it 3-inches in total
On day-4 he gives 4-inch piece and takes back the 1-inch and 2-inch pieces
On day-5 he gives 1-inch piece in addition to the already given 4-inch piece, thus making it 5-inches in total
This will continue till day-31.
Hence a minimum of five pieces is enough
Math logic:
Any number up to 2n-1 can be represented by the combinations of the 'n' numbers: 2^0,2^1,2^2,...2n-1.
For example, if 7 is considered, all numbers from 1 to 7 can be represented as combinations of 1,2, and 4:
1 = 1; 2 = 2; 3 = 1+2; 4 = 4; 5 = 1+4; 6 = 2+4; 7 = 1+2+4

If 9 is considered, all numbers from 1 to 9 can be represented as combinations of 1,2,4 and 2:
1 = 1; 2 = 2; 3 = 1+2; 4 = 4; 5 = 1+4; 6 = 2+4; 7 = 1+2+4; 8 = 2+2+4; 9 = 1+2+2+4;

If 15 is considered, all numbers from 1 to 15 can be represented as combinations of 1,2,4 and 8:
1 = 1; 2 = 2; 3 = 1+2; 4 = 4; 5 = 1+4; 6 = 2+4; 7 = 1+2+4; 8 = 8; 9 = 1+8; 10 = 2+8;
11 = 1+2+8; 12 = 4+8; 13 = 1+4+8; 14 = 2+4+8; 15 = 1+2+4+8

Answer (A)