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Question Number 113372 by Aina Samuel Temidayo last updated on 13/Sep/20

$$2^{\mathrm{a}}+2^{\mathrm{b}}+2^{\mathrm{c}}+2^{\mathrm{d}}=57,\mathrm{find}\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}.$$$$\mathrm{a}\ne \mathrm{b}\ne \mathrm{c}\ne \mathrm{d}\mathrm{and}\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\mathrm{are}\mathrm{positive}$$$$\mathrm{integers}.$$

Commented by udaythool last updated on 13/Sep/20

$$...\mathrm{is}\mathrm{there}\mathrm{any}\mathrm{restrictions}\mathrm{on}$$$$a,b,c,d?$$$$$$$$\mathrm{If}a,b,c,\mathrm{and}d\mathrm{are}\mathrm{integers}\mathrm{then}$$$$\mathrm{it}\mathrm{has}\mathrm{a}\mathrm{unique}\mathrm{solution},$$$$\mathrm{otherwise}\mathrm{infinitely}\mathrm{many}$$$$\mathrm{solutions}...$$

Commented by Aziztisffola last updated on 13/Sep/20

$$\frac{57}{2}\Rightarrow \mathrm{q}=28\mathrm{\&}\mathrm{r}=1$$$$\frac{28}{2}\Rightarrow \mathrm{q}=14\mathrm{\&}\mathrm{r}=0$$$$\frac{14}{2}\Rightarrow \mathrm{q}=7\mathrm{\&}\mathrm{r}=0$$$$\frac{7}{2}\Rightarrow \mathrm{q}=3\mathrm{\&}\mathrm{r}=1$$$$\frac{3}{2}\Rightarrow \mathrm{q}=1\mathrm{\&}\mathrm{r}=1$$$$57={\left(111001\right)}_2=2^0+2^3+2^4+2^5$$$$\Rightarrow 0+3+4+5=12$$

Answered by bemath last updated on 12/Sep/20

$$2^5+2^4+2^3+2^0=57$$$$\Rightarrow a+b+c+d=12$$

Commented by Aina Samuel Temidayo last updated on 12/Sep/20

$$\mathrm{Yea}.$$

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