jmc

Question

Stacy has d dollars. She enters a mall with 10 shops and a lottery stall. First she goes to the lottery and her money is doubled, then she goes into the first shop and spends 1024 dollars. After that she alternates playing the lottery and getting her money doubled (Stacy always wins) then going into a new shop and spending 1024. When she comes out of the last shop she has no money left. What is the minimum possible value of d$?

Accepted Answer

After the first lottery and first shop, Stacy has

2 d - 1024

dollars. After the second lottery and second shop, Stacy has

2 (2 d - 1024) - 1024 = 2^{2} d - (1 + 2) 1024

dollars. After the third lottery and third shop, Stacy has

2 (2^{2} d - (1 + 2) 1024) - 1024 = 2^{3} d - (1 + 2 + 2^{2}) 1024

dollars.

More generally, after the

n

th lottery and

n

th shop, Stacy has

2^{n} d - (1 + 2 + 2^{2} + \dots + 2^{n - 1}) 1024 = 2^{n} d - 1024 (2^{n} - 1)

dollars. In particular, for

n = 10,

Stacy has

1024 d - 1024 (1023)

dollars, which is also 0. Hence,

d = 1023 .