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ju

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22 and 1 wrapper left

15 chocolates from 45 Rs. generates 15 wrappers => 5 more chocolates generates 5wrappers => 1 more chocolate + 2 wrappers generates 1 more wrapper => 3 wrappers=> 1 more chocolate 1 wrapper left

Nice to find a formula for the general case whereas there's 'n' instead of '45'...

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• 15 ( 45 / 3) remainder = 0
• 5 (15 / 3) remainder = 0
• 1 ( 5 / 3) remainder 2
• 1 (1 + 2 from above)

Total: 22

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its -: 15+5+1+1

22 Chocolates And 1 wrapper is left

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22 chocolates and 1 wrapper left

How about this?

<span class="spoiler"><sup class="spoiler_label"> spoiler</sup>\sum_{i \mathop =0}^{∞}n/3^i</span>

good one just needs patience.

Actually, he can only buy 15 chocolates...the others are free.

A ruby solution. You could also build in different wrapper trade in ratios or different costs of chocolate bars.

class Buyer attrreader :chocbars def initialize(cash = 45, choc_bars = 0)

@cash = cash
@choc_bars = choc_bars

end

def getdatchocolate

spend_monies
trade_in_wrappers

end

private def tradeinwrappers

wrappers = @choc_bars
until wrappers <= 2
  # 3 for 1 more bar w/wrapper: 3 - 1 = 2
  wrappers -= 2
  @choc_bars += 1
end
@choc_bars

end

def spend_monies

@choc_bars += (@cash / 3)
@cash -= (@cash % 3)

end end

x = Buyer.new x.getdatchocolate

The real answer should be 15... He can only BUY 15.. the last 7 he got free for trading in wrappers.

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He can only BUY 15. The other 7 he gets are free as per the parameters of the question.

Rs. 45 = Rs.3 * 15 Chocolates 15 Chocolates = 15 Wrappers 3 Wrappers = 1 Chocolate 15 Wrappers = 5 Chocolates.. so totally 20 CHOCOLATES...!!!

Rs. 45 = Rs.3 * 15 Chocolates 15 Chocolates = 15 Wrappers 3 Wrappers = 1 Chocolate 15 Wrappers = 5 Chocolate(3 Wrappers = 1 Chocolates.)..

15+5+1+1=22 so totally 22 CHOCOLATES...!!!

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