The-sum-of-four-consecutive-2-digit-numbers-when-divided-by-10-becomes-a-perfect-square-Which-of-the-following-can-possibly-be-one-of-these-four-numbers-a-21-b-25-c-41-d-67-e-73-please-show-work

Question Number 19009 by chux last updated on 03/Aug/17

$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{four}\:\mathrm{consecutive} \\ $$$$\mathrm{2}−\mathrm{digit}\:\mathrm{numbers}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by} \\ $$$$\mathrm{10}\:\mathrm{becomes}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}.\mathrm{Which} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{can}\:\mathrm{possibly}\:\mathrm{be} \\ $$$$\mathrm{one}\:\mathrm{of}\:\mathrm{these}\:\mathrm{four}\:\mathrm{numbers}? \\ $$$$\left(\mathrm{a}\right)\mathrm{21}\left(\mathrm{b}\right)\mathrm{25}\left(\mathrm{c}\right)\mathrm{41}\left(\mathrm{d}\right)\mathrm{67}\left(\mathrm{e}\right)\mathrm{73} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{show}\:\mathrm{workings} \\ $$

Answered by ajfour last updated on 03/Aug/17

sum=10n^2 a=((5n^2 )/2)−(3/2) , b=((5n^2 )/2)−(1/2) c=((5n^2 )/2)+(1/2) , d=((5n^2 )/2)+(3/2) For n=3 , a=21 , so (a).

$$\mathrm{sum}=\mathrm{10n}^{\mathrm{2}} \\ $$$$\mathrm{a}=\frac{\mathrm{5n}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:,\:\:\:\mathrm{b}=\frac{\mathrm{5n}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{c}=\frac{\mathrm{5n}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:,\:\:\:\mathrm{d}=\frac{\mathrm{5n}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{For}\:\mathrm{n}=\mathrm{3}\:,\:\:\mathrm{a}=\mathrm{21}\:,\:\mathrm{so}\:\left(\mathrm{a}\right). \\ $$

Commented by chux last updated on 03/Aug/17

please how did you come about the 10n^2 and other representation

$$\mathrm{please}\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{come}\:\mathrm{about} \\ $$$$\mathrm{the}\:\mathrm{10n}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{other}\:\mathrm{representation} \\ $$

Commented by ajfour last updated on 03/Aug/17