Question Number 7249 by Rasheed Soomro last updated on 19/Aug/16
$${w},{x},{y},{z}\:{are}\:{digits}\:{in}\:{respective}\:{base}\:{system}\:{and} \\ $$$${a},{b}\:{are}\:{bases}. \\ $$$${Find}\:{out}\:{an}\:{example}/{examples}\:{which}\:{satisfy} \\ $$$${the}\:{following} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{wxyz}_{{a}} +{wxyz}_{{b}} ={wxyz}_{{a}+{b}} \: \\ $$
Commented by Yozzia last updated on 19/Aug/16
$$\mathrm{2}^{\mathrm{1}} +\mathrm{2}^{\mathrm{1}} =\left(\mathrm{2}+\mathrm{2}\right)^{\mathrm{1}} =\mathrm{4}^{\mathrm{1}} \\ $$$$\left\{\mathrm{1}×\mathrm{2}+\mathrm{0}×\mathrm{2}^{\mathrm{0}} \right\}+\left\{\mathrm{1}×\mathrm{2}+\mathrm{0}×\mathrm{2}^{\mathrm{0}} \right\}=\mathrm{1}×\mathrm{4}+\mathrm{0}×\mathrm{4}^{\mathrm{0}} \\ $$$${or}\:\left(\mathrm{10}\right)_{\mathrm{2}} +\left(\mathrm{10}\right)_{\mathrm{2}} =\left(\mathrm{10}\right)_{\mathrm{4}} \\ $$
Commented by Rasheed Soomro last updated on 19/Aug/16
$${Four}\:{digit}\:{number}\:{is}\:{required}. \\ $$
Commented by Yozzia last updated on 19/Aug/16
$${wxyz}_{{a}} +{wxyz}_{{b}} ={wxyz}_{{a}+{b}} \:\:\:\:\left(\ast\right) \\ $$$${Suppose}\:{wxyz}\:{is}\:{an}\:{integer}\:{identical}\:{on}\:{both}\:{sides}\:{of}\:\left(\ast\right) \\ $$$${e}.{g}\:\:\mathrm{1234}_{{a}} +\mathrm{1234}_{{b}} =\mathrm{1234}_{{a}+{b}} \: \\ $$$${for}\:{some}\:{given}\:{a},{b}\in\mathbb{N}. \\ $$$$\left[{wa}^{\mathrm{3}} +{xa}^{\mathrm{2}} +{ya}+{z}\right]+\left[{wb}^{\mathrm{3}} +{xb}^{\mathrm{2}} +{yb}+{z}\right]={w}\left({a}+{b}\right)^{\mathrm{3}} +{x}\left({a}+{b}\right)^{\mathrm{2}} +{y}\left({a}+{b}\right)+{z} \\ $$$${w}\left\{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} −\left({a}+{b}\right)^{\mathrm{3}} \right\}+{x}\left\{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} \right\}+{y}\left({a}+{b}−{a}−{b}\right)+\mathrm{2}{z}−{z}=\mathrm{0} \\ $$$${w}\left\{−\mathrm{3}{a}^{\mathrm{2}} {b}−\mathrm{3}{ab}^{\mathrm{2}} \right\}+{x}\left\{−\mathrm{2}{ab}\right\}+{z}=\mathrm{0} \\ $$$${w}\left\{\mathrm{3}{a}^{\mathrm{2}} {b}+\mathrm{3}{ab}^{\mathrm{2}} \right\}+\mathrm{2}{xab}−{z}=\mathrm{0}\:\: \\ $$$${The}\:{form}\:{of}\:{the}\:{above}\:{equation}\:{is}\: \\ $$$${independent}\:{of}\:{the}\:{value}\:{of}\:{y}\in\mathbb{N}. \\ $$$${z}={ab}\left(\mathrm{2}{x}+\mathrm{3}{w}\left({a}+{b}\right)\right) \\ $$$${w},{x}\in\mathbb{Z}^{\geqslant} ,\:{a},{b}\in\mathbb{N}−\left\{\mathrm{1}\right\}\:{and}\:{ab}\mid{z}.\:{If}\:{x}>\mathrm{0}\:{or}\:{w}>\mathrm{0} \\ $$$${then}\:\mathrm{2}{x}+\mathrm{3}{w}\left({a}+{b}\right)>\mathrm{1}\Rightarrow{z}>{ab}>{a}\:{and}\:{z}>{b}. \\ $$$${However},\:\mathrm{0}\leqslant{z}\leqslant{min}\left({a}−\mathrm{1},{b}−\mathrm{1}\right)<{min}\left({a},{b}\right) \\ $$$${for}\:{z}\:{in}\:{wxyz}_{{a}} \:{and}\:{wxyz}_{{b}} ,\:{a}\:{contradiction}. \\ $$$${Hence},\:{we}\:{must}\:{have}\:{x}={w}=\mathrm{0}\:\Rightarrow\:{z}=\mathrm{0}. \\ $$$$\therefore\:{we}\:{get}\:\mathrm{00}{y}\mathrm{0}_{{a}} +\mathrm{00}{y}\mathrm{0}_{{b}} =\mathrm{00}{y}\mathrm{0}_{{a}+{b}} \:\:{or}\:{y}\mathrm{0}_{{a}} +{y}\mathrm{0}_{{b}} ={y}\mathrm{0}_{{a}+{b}} \\ $$$${or}\:{ya}+{yb}={y}\left({a}+{b}\right)\:{for}\:{any}\:{y}\in\mathbb{N}\:,\mathrm{0}\leqslant{y}<{a},{y}<{b} \\ $$$${when}\:{a}\:{and}\:{b}\:{are}\:{known}\:{bases}. \\ $$$${No}\:{four}\:{digit}\:{integer}\:{wxyz}\:{satisfies} \\ $$$${wxyz}_{{a}} +{wxyz}_{{b}} ={wxyz}_{{a}+{b}} . \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 20/Aug/16
$$\mathcal{V}.\:\mathcal{N}{ic}\mathcal{E}! \\ $$$${What}\:{is}\:{meant}\:{by}\::\:\:\:\:\:{w},{x}\in\mathbb{Z}^{\geqslant} \:\:\:? \\ $$
Commented by Yozzia last updated on 20/Aug/16
$$\mathbb{Z}^{\geqslant} =\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},…\right\} \\ $$$${w},{x}\in\mathbb{Z}^{\geqslant} \equiv{x},{w}\:{are}\:{non}−{negative}\:{integers}. \\ $$
Commented by Rasheed Soomro last updated on 20/Aug/16
$$\mathcal{T}{h}\alpha{nk}\mathcal{S}! \\ $$
Commented by Yozzia last updated on 20/Aug/16