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Question Number 113134 by bemath last updated on 11/Sep/20

any one can explain me how to change decimal number to biner number. i′m forgot. example (315)_(10) = (...)_2 thank you

$$\mathrm{any}\:\mathrm{one}\:\mathrm{can}\:\mathrm{explain}\:\mathrm{me}\:\mathrm{how}\:\mathrm{to} \\ $$$$\mathrm{change}\:\mathrm{decimal}\:\mathrm{number}\:\mathrm{to}\: \\ $$$$\mathrm{biner}\:\mathrm{number}.\:\mathrm{i}'\mathrm{m}\:\mathrm{forgot}. \\ $$$$\mathrm{example}\:\left(\mathrm{315}\right)_{\mathrm{10}} \:=\:\left(...\right)_{\mathrm{2}} \\ $$$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by I want to learn more last updated on 11/Sep/20

2 315 Remainder −−−−∣−−−−−∣−−−−−−−− 2 ∣ 157 ∣ 1 2 ∣ 78 ∣ 1 2 ∣ 39 ∣ 0 2 ∣ 19 ∣ 1 2 ∣ 9 ∣ 1 2 ∣ 4 ∣ 1 2 ∣ 2 ∣ 0 2 ∣ 1 ∣ 0 2 ∣ 0 ∣ 1 read the remainder from bottom to top 315_(10) = 100111011_2

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{315}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Remainder} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−−−−\mid−−−−−\mid−−−−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\mathrm{157}\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\mathrm{78}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\mathrm{39}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\mathrm{19}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\mathrm{9}\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\mathrm{read}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{from}\:\mathrm{bottom}\:\mathrm{to}\:\mathrm{top} \\ $$$$\:\:\:\:\:\:\mathrm{315}_{\mathrm{10}} \:\:\:=\:\:\:\mathrm{100111011}_{\mathrm{2}} \\ $$

Commented by bemath last updated on 11/Sep/20

thank you friend

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{friend} \\ $$

Answered by MJS_new last updated on 11/Sep/20

find the largest n with 2^n ≤315 ⇒ n=8 (315)_(10) =(1a_7 a_6 a_5 a_4 a_3 a_2 a_1 a_0 )_2 315−2^8 =59 find the largest n with 2^n ≤59 ⇒ n=5 (315)_(10) =(1001a_4 a_3 a_2 a_1 a_0 )_2 59−2^5 =27 ...

$$\mathrm{find}\:\mathrm{the}\:\mathrm{largest}\:{n}\:\mathrm{with}\:\mathrm{2}^{{n}} \leqslant\mathrm{315} \\ $$$$\Rightarrow\:{n}=\mathrm{8} \\ $$$$\left(\mathrm{315}\right)_{\mathrm{10}} =\left(\mathrm{1}\underset{\mathrm{7}} {{a}}\underset{\mathrm{6}} {{a}}\underset{\mathrm{5}} {{a}}\underset{\mathrm{4}} {{a}}\underset{\mathrm{3}} {{a}}\underset{\mathrm{2}} {{a}}\underset{\mathrm{1}} {{a}}\underset{\mathrm{0}} {{a}}\right)_{\mathrm{2}} \\ $$$$\mathrm{315}−\mathrm{2}^{\mathrm{8}} =\mathrm{59} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{largest}\:{n}\:\mathrm{with}\:\mathrm{2}^{{n}} \leqslant\mathrm{59} \\ $$$$\Rightarrow\:{n}=\mathrm{5} \\ $$$$\left(\mathrm{315}\right)_{\mathrm{10}} =\left(\mathrm{1001}\underset{\mathrm{4}} {{a}}\underset{\mathrm{3}} {{a}}\underset{\mathrm{2}} {{a}}\underset{\mathrm{1}} {{a}}\underset{\mathrm{0}} {{a}}\right)_{\mathrm{2}} \\ $$$$\mathrm{59}−\mathrm{2}^{\mathrm{5}} =\mathrm{27} \\ $$$$... \\ $$

Commented by bemath last updated on 11/Sep/20

thank you prof

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{prof} \\ $$

Answered by mr W last updated on 11/Sep/20

⌊log_2 315⌋=8 315−2^8 =59 ⌊log_2 59⌋=5 59−2^5 =27 ⌊log_2 27⌋=4 27−2^4 =11 ⌊log_2 11⌋=3 11−2^3 =3 ⌊log_2 3⌋=1 3−2^1 =1=2^0 ⇒315=2^8 +2^5 +2^4 +2^3 +2^1 +2^0 =1×2^8 +0×2^7 +0×2^6 +1×2^5 +1×2^4 +1×2^3 +0×2^2 +1×2^1 +1×2^0 =(100111011)_2

$$\lfloor\mathrm{log}_{\mathrm{2}} \:\mathrm{315}\rfloor=\mathrm{8} \\ $$$$\mathrm{315}−\mathrm{2}^{\mathrm{8}} =\mathrm{59} \\ $$$$\lfloor\mathrm{log}_{\mathrm{2}} \:\mathrm{59}\rfloor=\mathrm{5} \\ $$$$\mathrm{59}−\mathrm{2}^{\mathrm{5}} =\mathrm{27} \\ $$$$\lfloor\mathrm{log}_{\mathrm{2}} \:\mathrm{27}\rfloor=\mathrm{4} \\ $$$$\mathrm{27}−\mathrm{2}^{\mathrm{4}} =\mathrm{11} \\ $$$$\lfloor\mathrm{log}_{\mathrm{2}} \:\mathrm{11}\rfloor=\mathrm{3} \\ $$$$\mathrm{11}−\mathrm{2}^{\mathrm{3}} =\mathrm{3} \\ $$$$\lfloor\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\rfloor=\mathrm{1} \\ $$$$\mathrm{3}−\mathrm{2}^{\mathrm{1}} =\mathrm{1}=\mathrm{2}^{\mathrm{0}} \\ $$$$\Rightarrow\mathrm{315}=\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{5}} +\mathrm{2}^{\mathrm{4}} +\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{1}} +\mathrm{2}^{\mathrm{0}} \\ $$$$=\mathrm{1}×\mathrm{2}^{\mathrm{8}} +\mathrm{0}×\mathrm{2}^{\mathrm{7}} +\mathrm{0}×\mathrm{2}^{\mathrm{6}} +\mathrm{1}×\mathrm{2}^{\mathrm{5}} +\mathrm{1}×\mathrm{2}^{\mathrm{4}} +\mathrm{1}×\mathrm{2}^{\mathrm{3}} +\mathrm{0}×\mathrm{2}^{\mathrm{2}} +\mathrm{1}×\mathrm{2}^{\mathrm{1}} +\mathrm{1}×\mathrm{2}^{\mathrm{0}} \\ $$$$=\left(\mathrm{100111011}\right)_{\mathrm{2}} \\ $$

Commented by bemath last updated on 11/Sep/20

thank you prof

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{prof} \\ $$

Answered by Aziztisffola last updated on 11/Sep/20

Commented by bemath last updated on 11/Sep/20

thank you friend

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{friend} \\ $$

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