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2-a-2-b-2-c-2-d-57-find-a-b-c-d-a-b-c-d-and-a-b-c-d-are-positive-integers-




Question Number 113372 by Aina Samuel Temidayo last updated on 13/Sep/20
2^a +2^b +2^c +2^d =57, find a+b+c+d.  a≠b≠c≠d and a,b,c,d are positive  integers.
$$\mathrm{2}^{\mathrm{a}} +\mathrm{2}^{\mathrm{b}} +\mathrm{2}^{\mathrm{c}} +\mathrm{2}^{\mathrm{d}} =\mathrm{57},\:\mathrm{find}\:\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}. \\ $$$$\mathrm{a}\neq\mathrm{b}\neq\mathrm{c}\neq\mathrm{d}\:\mathrm{and}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{are}\:\mathrm{positive} \\ $$$$\mathrm{integers}. \\ $$
Commented by udaythool last updated on 13/Sep/20
...is there any restrictions on  a, b, c, d?    If  a, b, c, and d are integers then  it has a unique solution,  otherwise infinitely many  solutions...
$$…\mathrm{is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{restrictions}\:\mathrm{on} \\ $$$${a},\:{b},\:{c},\:{d}? \\ $$$$ \\ $$$$\mathrm{If}\:\:{a},\:{b},\:{c},\:\mathrm{and}\:{d}\:\mathrm{are}\:\mathrm{integers}\:\mathrm{then} \\ $$$$\mathrm{it}\:\mathrm{has}\:\mathrm{a}\:\mathrm{unique}\:\mathrm{solution}, \\ $$$$\mathrm{otherwise}\:\mathrm{infinitely}\:\mathrm{many} \\ $$$$\mathrm{solutions}… \\ $$
Commented by Aziztisffola last updated on 13/Sep/20
((57)/2)⇒ q=28 & r=1  ((28)/2)⇒q=14 & r=0  ((14)/2)⇒q=7 & r=0  (7/2)⇒q=3 & r=1  (3/2)⇒q=1 &r=1  57=(111001)_2 =2^0 +2^3 +2^4 +2^5   ⇒0+3+4+5=12
$$\frac{\mathrm{57}}{\mathrm{2}}\Rightarrow\:\mathrm{q}=\mathrm{28}\:\&\:\mathrm{r}=\mathrm{1} \\ $$$$\frac{\mathrm{28}}{\mathrm{2}}\Rightarrow\mathrm{q}=\mathrm{14}\:\&\:\mathrm{r}=\mathrm{0} \\ $$$$\frac{\mathrm{14}}{\mathrm{2}}\Rightarrow\mathrm{q}=\mathrm{7}\:\&\:\mathrm{r}=\mathrm{0} \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}\Rightarrow\mathrm{q}=\mathrm{3}\:\&\:\mathrm{r}=\mathrm{1} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow\mathrm{q}=\mathrm{1}\:\&\mathrm{r}=\mathrm{1} \\ $$$$\mathrm{57}=\left(\mathrm{111001}\right)_{\mathrm{2}} =\mathrm{2}^{\mathrm{0}} +\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{4}} +\mathrm{2}^{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{0}+\mathrm{3}+\mathrm{4}+\mathrm{5}=\mathrm{12} \\ $$
Answered by bemath last updated on 12/Sep/20
2^5 +2^4 +2^3 +2^0  = 57  ⇒a+b+c+d=12
$$\mathrm{2}^{\mathrm{5}} +\mathrm{2}^{\mathrm{4}} +\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{0}} \:=\:\mathrm{57} \\ $$$$\Rightarrow{a}+{b}+{c}+{d}=\mathrm{12} \\ $$
Commented by Aina Samuel Temidayo last updated on 12/Sep/20
Yea.
$$\mathrm{Yea}. \\ $$

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