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Question Number 55674 by gunawan last updated on 01/Mar/19

The smallest integer numbers with n ≥ 2018 so ((√3)+3i)^n form real numbers is..

$$\mathrm{The}\:\mathrm{smallest}\:\mathrm{integer}\:\mathrm{numbers} \\ $$$$\mathrm{with}\:{n}\:\geqslant\:\mathrm{2018}\:\mathrm{so} \\ $$$$\left(\sqrt{\mathrm{3}}+\mathrm{3}{i}\right)^{{n}} \:\mathrm{form}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{is}.. \\ $$

Answered by mr W last updated on 02/Mar/19

((√3)+3i)^n =(2(√3))^n ((1/2)+((√3)/2)i)^n =(2(√3))^n (cos (π/3)+i sin (π/3))^n =(2(√3))^n (cos ((nπ)/3)+i sin ((nπ)/3)) ⇒sin ((nπ)/3)=0 ⇒((nπ)/3)=mπ ⇒n=3m≥2018 ⇒n_(min) =2019

$$\left(\sqrt{\mathrm{3}}+\mathrm{3}{i}\right)^{{n}} \\ $$$$=\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right)^{{n}} \\ $$$$=\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{{n}} \left(\mathrm{cos}\:\frac{\pi}{\mathrm{3}}+{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\right)^{{n}} \\ $$$$=\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{{n}} \left(\mathrm{cos}\:\frac{{n}\pi}{\mathrm{3}}+{i}\:\mathrm{sin}\:\frac{{n}\pi}{\mathrm{3}}\right) \\ $$$$\Rightarrow\mathrm{sin}\:\frac{{n}\pi}{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{n}\pi}{\mathrm{3}}={m}\pi \\ $$$$\Rightarrow{n}=\mathrm{3}{m}\geqslant\mathrm{2018} \\ $$$$\Rightarrow{n}_{{min}} =\mathrm{2019} \\ $$

Commented by gunawan last updated on 02/Mar/19

Nice thank you Sir

$$\mathrm{Nice}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

Commented by malwaan last updated on 02/Mar/19

what is the point ? we can write 2019 without proof please explane sir

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{point}\:? \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{write}\:\mathrm{2019} \\ $$$$\mathrm{without}\:\mathrm{proof}\: \\ $$$$\mathrm{please}\:\mathrm{explane}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 02/Mar/19

⇒n=3m≥2018 means n must be divisible by 3 and at least 2018. a number is divisible by 3 if the sum of its digits is divisible by 3. with 2018 the sum of its digits is 2+0+1+8=11, not divisible by 3, but 2+0+1+9=12, it′s divisible by 3, therefore n_(min) =2019.

$$\Rightarrow{n}=\mathrm{3}{m}\geqslant\mathrm{2018} \\ $$$${means}\:{n}\:{must}\:{be}\:{divisible}\:{by}\:\mathrm{3}\:{and} \\ $$$${at}\:{least}\:\mathrm{2018}. \\ $$$$ \\ $$$${a}\:{number}\:{is}\:{divisible}\:{by}\:\mathrm{3}\:{if}\:{the} \\ $$$${sum}\:{of}\:{its}\:{digits}\:{is}\:{divisible}\:{by}\:\mathrm{3}. \\ $$$$ \\ $$$${with}\:\mathrm{2018}\:{the}\:{sum}\:{of}\:{its}\:{digits}\:{is}\: \\ $$$$\mathrm{2}+\mathrm{0}+\mathrm{1}+\mathrm{8}=\mathrm{11},\:{not}\:{divisible}\:{by}\:\mathrm{3}, \\ $$$${but}\:\mathrm{2}+\mathrm{0}+\mathrm{1}+\mathrm{9}=\mathrm{12},\:{it}'{s}\:{divisible}\:{by}\:\mathrm{3}, \\ $$$${therefore}\:{n}_{{min}} =\mathrm{2019}. \\ $$

Commented by malwaan last updated on 03/Mar/19

thank you very much sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$

Commented by Kunal12588 last updated on 03/Mar/19

sir i heard this for checking divisibility with 3 we have to add the digits of the number special case:− if a digit contains 9 we can drop the 9 and add other digits eg:− 2019 − 2+0+1=3 which is divisible by 3. why is this sir?

$${sir}\:{i}\:{heard}\:{this}\: \\ $$$${for}\:{checking}\:{divisibility}\:{with}\:\mathrm{3} \\ $$$${we}\:{have}\:{to}\:{add}\:{the}\:{digits}\:{of}\:{the}\:{number} \\ $$$${special}\:{case}:−\:{if}\:{a}\:{digit}\:{contains}\:\mathrm{9} \\ $$$${we}\:{can}\:{drop}\:{the}\:\mathrm{9}\:{and}\:{add}\:{other}\:{digits} \\ $$$${eg}:−\:\mathrm{2019}\:−\:\mathrm{2}+\mathrm{0}+\mathrm{1}=\mathrm{3}\:{which}\:{is}\:{divisible}\:{by}\:\mathrm{3}. \\ $$$${why}\:{is}\:{this}\:{sir}? \\ $$

Commented by mr W last updated on 03/Mar/19

because digit 9 is already divisibly by 3. if the sum of other digits is divisible by 3, then the total sum of all digits is also divisible by 3 and the number is divisible by 3. in fact you can drop not only digit 9, but also digits 6, 3 and 0, and only need to add other remaining digits. e.g. we take the number 123450906497628005 we only need to add the blue digits: 1+2+4+5+4+7+2+8+5=38 which is not divisible by 3, therefore the given number is not divisible by 3.

$${because}\:{digit}\:\mathrm{9}\:{is}\:{already}\:{divisibly}\:{by}\:\mathrm{3}. \\ $$$${if}\:{the}\:{sum}\:{of}\:{other}\:{digits}\:{is}\:{divisible} \\ $$$${by}\:\mathrm{3},\:{then}\:{the}\:{total}\:{sum}\:{of}\:{all}\:{digits}\:{is} \\ $$$${also}\:{divisible}\:{by}\:\mathrm{3}\:{and}\:{the}\:{number}\:{is} \\ $$$${divisible}\:{by}\:\mathrm{3}.\:\:{in}\:{fact}\:{you}\:{can}\:{drop}\:{not} \\ $$$${only}\:{digit}\:\mathrm{9},\:{but}\:{also}\:{digits}\:\mathrm{6},\:\mathrm{3}\:{and}\:\mathrm{0}, \\ $$$${and}\:{only}\:{need}\:{to}\:{add}\:{other}\:{remaining} \\ $$$${digits}. \\ $$$${e}.{g}.\:{we}\:{take}\:{the}\:{number} \\ $$$$\mathrm{123450906497628005} \\ $$$${we}\:{only}\:{need}\:{to}\:{add}\:{the}\:{blue}\:{digits}: \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{5}+\mathrm{4}+\mathrm{7}+\mathrm{2}+\mathrm{8}+\mathrm{5}=\mathrm{38}\:{which} \\ $$$${is}\:{not}\:{divisible}\:{by}\:\mathrm{3},\:{therefore}\:{the} \\ $$$${given}\:{number}\:{is}\:{not}\:{divisible}\:{by}\:\mathrm{3}. \\ $$

Commented by Kunal12588 last updated on 03/Mar/19

great sir. and thanks

$${great}\:{sir}.\:{and}\:{thanks} \\ $$

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