Hi,
Came across an interesting puzzle..
was trying to find solution via Excel...
You get three candies for 1$.And if you exchange three candy covers,you get an extra candy.
So how many candies will you get in total for a price of 45$?
Regards
Hi,
Came across an interesting puzzle..
was trying to find solution via Excel...
You get three candies for 1$.And if you exchange three candy covers,you get an extra candy.
So how many candies will you get in total for a price of 45$?
Regards
45$ buys 135 bars
trade the above at 3 wrappers per candy we get 45candies
trade again we get 15 candies
and again we get 5 candies
and again we get 1 candy + 2 spare wrappers
so we get a final 1 candy
=202
????
Last edited by JohnTopley; 01-07-2017 at 10:33 AM.
Hi John,
Its the right answer!!
However I was thinking if its possible to get the answer with the help of a formula
I am no mathematician ...
but we start with 5 and work "upwards" to 135 we have a geometric progression
=5* (1-3^4)/(1-3)
r=ratio : 3 in this case and 4 terms
this gives a result of 200
I don't know how you add the 1s (2 of them)!
You can use this formula in excel.
Where J5 has dollarsPlease Login or Register to view this content.
Thanks
Mark the thread as solved if you are satisfied with the answer.
In your first post under the thread tools.
Mahju
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