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Question-195747




Question Number 195747 by Mastermind last updated on 09/Aug/23
Commented by AST last updated on 09/Aug/23
The total amount spent does not have to be  necessarily equal to the sum of the remaining  amount at each step of your spending. In fact,  we can make that N51 grow arbitrarily large  by spending ε(such as 0.000000001) at every step.
$${The}\:{total}\:{amount}\:{spent}\:{does}\:{not}\:{have}\:{to}\:{be} \\ $$$${necessarily}\:{equal}\:{to}\:{the}\:{sum}\:{of}\:{the}\:{remaining} \\ $$$${amount}\:{at}\:{each}\:{step}\:{of}\:{your}\:{spending}.\:{In}\:{fact}, \\ $$$${we}\:{can}\:{make}\:{that}\:{N}\mathrm{51}\:{grow}\:{arbitrarily}\:{large} \\ $$$${by}\:{spending}\:\epsilon\left({such}\:{as}\:\mathrm{0}.\mathrm{000000001}\right)\:{at}\:{every}\:{step}. \\ $$
Commented by Mastermind last updated on 09/Aug/23
Wow  thank you man
$$\mathrm{Wow} \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{man} \\ $$
Commented by Frix last updated on 09/Aug/23
It′s a bit like counting the fingers of one  hand backwards & adding the other hand:  Left hand: 10−9−8−7−6  Right hand: 5  6+5=11
$$\mathrm{It}'\mathrm{s}\:\mathrm{a}\:\mathrm{bit}\:\mathrm{like}\:\mathrm{counting}\:\mathrm{the}\:\mathrm{fingers}\:\mathrm{of}\:\mathrm{one} \\ $$$$\mathrm{hand}\:\mathrm{backwards}\:\&\:\mathrm{adding}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hand}: \\ $$$$\mathrm{Left}\:\mathrm{hand}:\:\mathrm{10}−\mathrm{9}−\mathrm{8}−\mathrm{7}−\mathrm{6} \\ $$$$\mathrm{Right}\:\mathrm{hand}:\:\mathrm{5} \\ $$$$\mathrm{6}+\mathrm{5}=\mathrm{11} \\ $$

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