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2-a-4-b-8-c-328-find-a-b-and-c-when-a-b-c-is-natual-number-

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Question Number 191663 by mathlove last updated on 28/Apr/23
2^a +4^b +8^c =328 find a,b and c when(a,b,c)is natual number
$$\mathrm{2}^{{a}} +\mathrm{4}^{{b}} +\mathrm{8}^{{c}} =\mathrm{328} \\ $$$${find}\:{a},{b}\:{and}\:{c} \\ $$$${when}\left({a},{b},{c}\right){is}\:{natual}\:{number} \\ $$
Answered by Rasheed.Sindhi last updated on 28/Apr/23
2^a +4^b +8^c =328 2^a +2^(2b) +2^(3c) =328=101001000_2 =2^8 +2^6 +2^3 3c=3,6⇒c=1,2 2b=6,8⇒b=3,4 a=3,6,8 c= { ((1⇒b= { ((3⇒a=8)),((4⇒a=6)) :})),((2⇒b=4⇒a=3)) :} (a,b,c)=(3,4,2),(6,4,1),(8,3,1)
$$\mathrm{2}^{{a}} +\mathrm{4}^{{b}} +\mathrm{8}^{{c}} =\mathrm{328} \\ $$$$\mathrm{2}^{{a}} +\mathrm{2}^{\mathrm{2}{b}} +\mathrm{2}^{\mathrm{3}{c}} =\mathrm{328}=\mathrm{101001000}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{6}} +\mathrm{2}^{\mathrm{3}} \\ $$$$\mathrm{3}{c}=\mathrm{3},\mathrm{6}\Rightarrow{c}=\mathrm{1},\mathrm{2} \\ $$$$\mathrm{2}{b}=\mathrm{6},\mathrm{8}\Rightarrow{b}=\mathrm{3},\mathrm{4} \\ $$$${a}=\mathrm{3},\mathrm{6},\mathrm{8} \\ $$$${c}=\begin{cases}{\mathrm{1}\Rightarrow{b}=\begin{cases}{\mathrm{3}\Rightarrow{a}=\mathrm{8}}\\{\mathrm{4}\Rightarrow{a}=\mathrm{6}}\end{cases}}\\{\mathrm{2}\Rightarrow{b}=\mathrm{4}\Rightarrow{a}=\mathrm{3}}\end{cases} \\ $$$$\left({a},{b},{c}\right)=\left(\mathrm{3},\mathrm{4},\mathrm{2}\right),\left(\mathrm{6},\mathrm{4},\mathrm{1}\right),\left(\mathrm{8},\mathrm{3},\mathrm{1}\right) \\ $$$$ \\ $$
Commented by BaliramKumar last updated on 28/Apr/23
Nice solution Sir
$$\mathrm{Nice}\:\mathrm{solution}\:\mathrm{Sir} \\ $$
Commented by mathlove last updated on 28/Apr/23
thanks sir
$${thanks}\:{sir} \\ $$
Answered by BaliramKumar last updated on 28/Apr/23
2^a +2^(2b) +2^(3c) = 2^3 ×41 2^(3c) (2^(a−3c) +2^(2b−3c) +1) = 2^3 ×41 c = 1 2^(a−3) +2^(2b−3) +1 = 41 2^(a−3) +2^(2b−3) = 40 2^(a−3) +2^(2b−3) = 2^3 ×5 2^a +2^(2b) = 2^6 ×5 2^(2b) (2^(a−2b) +1) = 2^6 ×5 b = 3 2^(a−2×3) +1 = 5 2^(a−6) = 5−1 = 4 = 2^2 a−6 = 2 a = 8
$$\mathrm{2}^{{a}} +\mathrm{2}^{\mathrm{2}{b}} +\mathrm{2}^{\mathrm{3}{c}} \:=\:\mathrm{2}^{\mathrm{3}} ×\mathrm{41} \\ $$$$\mathrm{2}^{\mathrm{3}{c}} \left(\mathrm{2}^{{a}−\mathrm{3}{c}} +\mathrm{2}^{\mathrm{2}{b}−\mathrm{3}{c}} +\mathrm{1}\right)\:=\:\mathrm{2}^{\mathrm{3}} ×\mathrm{41} \\ $$$${c}\:=\:\mathrm{1} \\ $$$$\mathrm{2}^{{a}−\mathrm{3}} +\mathrm{2}^{\mathrm{2}{b}−\mathrm{3}} +\mathrm{1}\:=\:\mathrm{41} \\ $$$$\mathrm{2}^{{a}−\mathrm{3}} +\mathrm{2}^{\mathrm{2}{b}−\mathrm{3}} \:=\:\mathrm{40} \\ $$$$\mathrm{2}^{{a}−\mathrm{3}} +\mathrm{2}^{\mathrm{2}{b}−\mathrm{3}} \:=\:\mathrm{2}^{\mathrm{3}} ×\mathrm{5} \\ $$$$\mathrm{2}^{{a}} +\mathrm{2}^{\mathrm{2}{b}} \:=\:\mathrm{2}^{\mathrm{6}} ×\mathrm{5} \\ $$$$\mathrm{2}^{\mathrm{2}{b}} \left(\mathrm{2}^{{a}−\mathrm{2}{b}} +\mathrm{1}\right)\:=\:\mathrm{2}^{\mathrm{6}} ×\mathrm{5} \\ $$$${b}\:=\:\mathrm{3} \\ $$$$\mathrm{2}^{{a}−\mathrm{2}×\mathrm{3}} +\mathrm{1}\:=\:\mathrm{5} \\ $$$$\mathrm{2}^{{a}−\mathrm{6}} \:=\:\mathrm{5}−\mathrm{1}\:=\:\mathrm{4}\:=\:\mathrm{2}^{\mathrm{2}} \\ $$$${a}−\mathrm{6}\:=\:\mathrm{2} \\ $$$${a}\:=\:\mathrm{8} \\ $$
Commented by mathlove last updated on 28/Apr/23
thanks
$${thanks} \\ $$
Answered by mehdee42 last updated on 28/Apr/23
number 328 is made by combining the sum of three numbers 256,64 & 8 .according to the experssions in the given eqution ,we hsve the following three states. (2^a ,4^b ,8^c )=(256,64,8)=(64,256,8)=(8,256,64) ⇒(a,b,c)=(8,3,1)∨(6,4,1)∨(3,4,2)
$${number}\:\mathrm{328}\:{is}\:{made}\:{by}\:{combining}\:{the}\:{sum}\:{of}\:{three}\:{numbers}\:\mathrm{256},\mathrm{64}\:\&\:\mathrm{8}\:.{according}\:{to}\:{the}\:{experssions}\:{in}\:{the}\:{given}\:{eqution}\:,{we}\:{hsve}\:{the}\:{following}\:\:{three}\:{states}. \\ $$$$\left(\mathrm{2}^{{a}} ,\mathrm{4}^{{b}} ,\mathrm{8}^{{c}} \right)=\left(\mathrm{256},\mathrm{64},\mathrm{8}\right)=\left(\mathrm{64},\mathrm{256},\mathrm{8}\right)=\left(\mathrm{8},\mathrm{256},\mathrm{64}\right) \\ $$$$\Rightarrow\left({a},{b},{c}\right)=\left(\mathrm{8},\mathrm{3},\mathrm{1}\right)\vee\left(\mathrm{6},\mathrm{4},\mathrm{1}\right)\vee\left(\mathrm{3},\mathrm{4},\mathrm{2}\right)\: \\ $$
Answered by LOSER last updated on 29/Apr/23
2^a ; 4^b ; 8^c > 0 ∀ a; b; c ∈ R ⇒ 8^c < 328 ⇒ c = 2; c = 1; c = 0 • c = 2 ⇒ 2^a + 4^b = 264 ⇒ 4^b < 264 ⇒ b = 4; b = 3; b = 2; b = 1; b = 0 ∗ b = 4 ⇒ a = 3 ∗ b = 3; b = 2; b = 1; b = 0 ⇒ ∄ a • c = 1 ⇒ 2^a + 4^b = 320 ⇒ 4^b < 320 ⇒ b = 4; b = 3; b = 2; b = 1; b = 0 ∗ b = 4 ⇒ a = 6 ∗ b = 3 ⇒ a = 8 ∗ b = 2; b = 1; b = 0 ⇒ ∄ a • c = 0 ⇒ 2^a + 4^b = 327 ⇒ b = 4; b = 3; b = 2; b = 1; b = 0 ⇒ ∄ a ⇒ (a; b; c) = (3; 4; 2); (6; 4; 1); (8; 3; 1)
$$\mathrm{2}^{{a}} ;\:\mathrm{4}^{{b}} ;\:\mathrm{8}^{{c}} \:>\:\mathrm{0}\:\forall\:{a};\:{b};\:{c}\:\in\:\mathbb{R} \\ $$$$\Rightarrow\:\mathrm{8}^{{c}} \:<\:\mathrm{328}\:\Rightarrow\:{c}\:=\:\mathrm{2};\:{c}\:=\:\mathrm{1};\:{c}\:=\:\mathrm{0} \\ $$$$\bullet\:{c}\:=\:\mathrm{2}\: \\ $$$$\Rightarrow\:\mathrm{2}^{{a}} \:+\:\mathrm{4}^{{b}} \:=\:\mathrm{264}\:\Rightarrow\:\mathrm{4}^{{b}} \:<\:\mathrm{264} \\ $$$$\Rightarrow\:{b}\:=\:\mathrm{4};\:{b}\:=\:\mathrm{3};\:{b}\:=\:\mathrm{2};\:{b}\:=\:\mathrm{1};\:{b}\:=\:\mathrm{0} \\ $$$$\ast\:{b}\:=\:\mathrm{4}\:\Rightarrow\:{a}\:=\:\mathrm{3} \\ $$$$\ast\:{b}\:=\:\mathrm{3};\:{b}\:=\:\mathrm{2};\:{b}\:=\:\mathrm{1};\:{b}\:=\:\mathrm{0}\:\Rightarrow\:\nexists\:{a} \\ $$$$\bullet\:{c}\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{2}^{{a}} \:+\:\mathrm{4}^{{b}} \:=\:\mathrm{320}\:\Rightarrow\:\mathrm{4}^{{b}} \:<\:\mathrm{320} \\ $$$$\Rightarrow\:{b}\:=\:\mathrm{4};\:{b}\:=\:\mathrm{3};\:{b}\:=\:\mathrm{2};\:{b}\:=\:\mathrm{1};\:{b}\:=\:\mathrm{0} \\ $$$$\ast\:{b}\:=\:\mathrm{4}\:\Rightarrow\:{a}\:=\:\mathrm{6} \\ $$$$\ast\:{b}\:=\:\mathrm{3}\:\Rightarrow\:{a}\:=\:\mathrm{8} \\ $$$$\ast\:{b}\:=\:\mathrm{2};\:{b}\:=\:\mathrm{1};\:{b}\:=\:\mathrm{0}\:\Rightarrow\:\nexists\:{a} \\ $$$$\bullet\:{c}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{2}^{{a}} \:+\:\mathrm{4}^{{b}} \:=\:\mathrm{327} \\ $$$$\Rightarrow\:{b}\:=\:\mathrm{4};\:{b}\:=\:\mathrm{3};\:{b}\:=\:\mathrm{2};\:{b}\:=\:\mathrm{1};\:{b}\:=\:\mathrm{0}\:\Rightarrow\:\nexists\:{a} \\ $$$$\Rightarrow\:\left({a};\:{b};\:{c}\right)\:=\:\left(\mathrm{3};\:\mathrm{4};\:\mathrm{2}\right);\:\left(\mathrm{6};\:\mathrm{4};\:\mathrm{1}\right);\:\left(\mathrm{8};\:\mathrm{3};\:\mathrm{1}\right) \\ $$

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