Question Number 21353 by Tinkutara last updated on 21/Sep/17
$$\mathrm{A}\:\mathrm{censusman}\:\mathrm{on}\:\mathrm{duty}\:\mathrm{visited}\:\mathrm{a}\:\mathrm{house} \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{lady}\:\mathrm{inmates}\:\mathrm{declined}\:\mathrm{to} \\ $$$$\mathrm{reveal}\:\mathrm{their}\:\mathrm{individual}\:\mathrm{ages},\:\mathrm{but}\:\mathrm{said}\:− \\ $$$$“\mathrm{we}\:\mathrm{do}\:\mathrm{not}\:\mathrm{mind}\:\mathrm{giving}\:\mathrm{you}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{ages}\:\mathrm{of}\:\mathrm{any}\:\mathrm{two}\:\mathrm{ladies}\:\mathrm{you}\:\mathrm{may} \\ $$$$\mathrm{choose}''.\:\mathrm{Thereupon}\:\mathrm{the}\:\mathrm{censusman} \\ $$$$\mathrm{said}\:−\:“\mathrm{In}\:\mathrm{that}\:\mathrm{case}\:\mathrm{please}\:\mathrm{give}\:\mathrm{me}\:\mathrm{the} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ages}\:\mathrm{of}\:\mathrm{every}\:\mathrm{possible}\:\mathrm{pair}\:\mathrm{of} \\ $$$$\mathrm{you}''.\:\mathrm{The}\:\mathrm{gave}\:\mathrm{the}\:\mathrm{sums}\:\mathrm{as}\:\mathrm{follows}\:: \\ $$$$\mathrm{30},\:\mathrm{33},\:\mathrm{41},\:\mathrm{58},\:\mathrm{66},\:\mathrm{69}.\:\mathrm{The}\:\mathrm{censusman} \\ $$$$\mathrm{took}\:\mathrm{these}\:\mathrm{figures}\:\mathrm{and}\:\mathrm{happily}\:\mathrm{went} \\ $$$$\mathrm{away}.\:\mathrm{How}\:\mathrm{did}\:\mathrm{he}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{individual} \\ $$$$\mathrm{ages}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ladies}\:\mathrm{from}\:\mathrm{these}\:\mathrm{figures}? \\ $$
Answered by dioph last updated on 21/Sep/17
![6 sums indicate there are 4 ladies with ages a < b < c < d. a+b < a+c < [(b+c), (a+d)] < b+d < c+d { ((a+b = 30)),((a+c = 33)),((b+d = 66)),((c+d = 69)) :} a=[(a+b)+(a+c)−(b+c)]/2 b+c = 58 ⇒ a = 5/2 ∉ Z hence b+c = 41 and a+d = 58 a = 11, b = 19, c = 22, d = 47](https://www.tinkutara.com/question/Q21365.png)
$$\mathrm{6}\:\mathrm{sums}\:\mathrm{indicate}\:\mathrm{there}\:\mathrm{are}\:\mathrm{4}\:\mathrm{ladies}\:\mathrm{with} \\ $$$$\mathrm{ages}\:{a}\:<\:{b}\:<\:{c}\:<\:{d}. \\ $$$${a}+{b}\:<\:{a}+{c}\:<\:\left[\left({b}+{c}\right),\:\left({a}+{d}\right)\right]\:<\:{b}+{d}\:<\:{c}+{d} \\ $$$$\begin{cases}{{a}+{b}\:=\:\mathrm{30}}\\{{a}+{c}\:=\:\mathrm{33}}\\{{b}+{d}\:=\:\mathrm{66}}\\{{c}+{d}\:=\:\mathrm{69}}\end{cases} \\ $$$${a}=\left[\left({a}+{b}\right)+\left({a}+{c}\right)−\left({b}+{c}\right)\right]/\mathrm{2} \\ $$$${b}+{c}\:=\:\mathrm{58}\:\Rightarrow\:{a}\:=\:\mathrm{5}/\mathrm{2}\:\notin\:\mathbb{Z} \\ $$$$\mathrm{hence}\:{b}+{c}\:=\:\mathrm{41}\:\mathrm{and}\:{a}+{d}\:=\:\mathrm{58} \\ $$$${a}\:=\:\mathrm{11},\:{b}\:=\:\mathrm{19},\:{c}\:=\:\mathrm{22},\:{d}\:=\:\mathrm{47} \\ $$
Commented by Tinkutara last updated on 22/Sep/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$