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

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

Answered by ajfour last updated on 03/Aug/17

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

Commented by chux last updated on 03/Aug/17

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

Commented by ajfour last updated on 03/Aug/17

$$\frac{\mathrm{sum}}{10}={\mathrm{n}}^2\left(\mathrm{the}\mathrm{perfect}\mathrm{square}\right)$$

Commented by chux last updated on 03/Aug/17

$$\mathrm{ok}.....\mathrm{its}\mathrm{clear}.$$

Commented by chernoaguero@gmail.com last updated on 03/Aug/17

$$siradfourwhyis\pm \frac{3}{2}and\pm \frac{1}{2}applicable$$

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