Work backward

Question

on Saturday Mr. Singh sold a lot of watermelons. His firtd customer bought 1 more then 1/2 of all the watermelons. The second customer bought 1 more than 1/2 of the watermelons that were left. The third, fourth and fifth customers each bought 1 more than 1/2 of the watermelons that were left after the customer before Them. Customer 6 bought only the 1 remaining watermelo. How many watermelons did Mr. Singh start with on Saturday?

Answer 1

Let x = number of watermelons he started with.

 

 

Then the first customer bought (1/2)x + 1 watermelons. So, afterwards, there were:

 

x - ((1/2x) + 1)

x - (1/2)x - 1

(1/2)x - 1 watermelons

 

 

The second customer bought 1/2((1/2)x - 1) + 1 watermelons. Afterwards there were:

 

(1/2)x - 1 - (1/2((1/2)x - 1) + 1)

(1/2)x - 1 - ((1/4)x - 1/2 + 1)

(1/2)x - 1 - ((1/4)x + 1/2)

(1/2)x - 1 - (1/4)x - 1/2

(1/4)x - 3/2 watermelons

 

 

The third customer bought 1/2((1/4)x - 3/2) + 1 watermelons. Afterwards there were:

 

(1/4)x - 3/2 - (1/2((1/4)x - 3/2) + 1)

(1/4)x - 3/2 - ((1/8)x - 3/4 + 1)

(1/4)x - 3/2 - ((1/8)x + 1/4)

(1/4)x - 3/2 - (1/8)x - 1/4

(1/8)x - 7/4 watermelons

 

 

Each time, the x term is being divided by 2. Each time, we subtract 1, then 1/2, then 1/4, then 1/8... from the constant term. Continuing the pattern, after the fourth customer there were:

 

(1/2)(1/8)x - 7/4 - 1/8

(1/16)x - 15/8 watermelons

 

 

After the fifth customer there were:

 

(1/2)(1/16)x - 15/8 - 1/16

(1/32)x - 31/16

 

 

The sixth customer buys only 1 watermelon, leaving 0 watermelons:

 

(1/32)x - 31/16 - 1 = 0

(1/32)x - 47/16 = 0

(1/32)x = 47/16

x = 32(47/16)

x = 2(47/1)

x = 94

 

 

The man started with 94 watermelons.

 

 

Check our work:

 

Customer 1 buys (1/2)(94) + 1 = 48, leaving 46 watermelons.

Customer 2 buys (1/2)(46) + 1 = 24, leaving 22 watermelons.

Customer 3 buys (1/2)(22) + 1 = 12, leaving 10 watermelons.

Customer 4 buys (1/2)(10) + 1 = 6, leaving 4 watermelons.

Customer 5 buys (1/2)(4) + 1 = 3, leaving 1 watermelon.

Customer 6 buys the only 1 remaining, leaving 0 watermelons.

 

That checks out!

 

 

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By the way, the pattern above, before we found an actual number, can be represented by this equation:

 

a_n = x/(2ⁿ) - (2ⁿ⁺¹ - 2)/(2ⁿ) = x/(2ⁿ) - 2 + 1/(2^n-1),

 

where n = the number of people who have bought watermelons and a_n = the number of watermelons remaining.