Question Number 217071 by ArshadS last updated on 28/Feb/25

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{two}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{that}\:\mathrm{are}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{four}\:\mathrm{times}\:\mathrm{the}\:\mathrm{sum}\: \\ $$$$\mathrm{of}\:\mathrm{their}\:\mathrm{digits}.\:\mathrm{Solve}\:\mathrm{this}\:\mathrm{using}\:\mathrm{at}\:\mathrm{least}\:\mathrm{two}\:\mathrm{different}\:\mathrm{methods}\: \\ $$$$\mathrm{and}\:\mathrm{verify}\:\mathrm{your}\:\mathrm{answers}. \\ $$
Answered by som(math1967) last updated on 28/Feb/25
![x+10y=4(x+y) ⇒3x=6y⇒x:y=2:1 12 [ 4×(1+2)=12] 24 [4×(2+4)=24] 36 [4×(3+6)=36] 48 [4×(4+8)=48]](https://www.tinkutara.com/question/Q217074.png)
$$\:{x}+\mathrm{10}{y}=\mathrm{4}\left({x}+{y}\right) \\ $$$$\Rightarrow\mathrm{3}{x}=\mathrm{6}{y}\Rightarrow{x}:{y}=\mathrm{2}:\mathrm{1} \\ $$$$\mathrm{12}\:\left[\:\mathrm{4}×\left(\mathrm{1}+\mathrm{2}\right)=\mathrm{12}\right] \\ $$$$\mathrm{24}\:\left[\mathrm{4}×\left(\mathrm{2}+\mathrm{4}\right)=\mathrm{24}\right] \\ $$$$\mathrm{36}\:\left[\mathrm{4}×\left(\mathrm{3}+\mathrm{6}\right)=\mathrm{36}\right] \\ $$$$\mathrm{48}\:\left[\mathrm{4}×\left(\mathrm{4}+\mathrm{8}\right)=\mathrm{48}\right] \\ $$$$\:\: \\ $$
Commented by ArshadS last updated on 28/Feb/25

$${nice}!\:{thanks}\:{sir}! \\ $$