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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..
Thesmallestintegernumberswithn2018so(3+3i)nformrealnumbersis..
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
(3+3i)n=(23)n(12+32i)n=(23)n(cosπ3+isinπ3)n=(23)n(cosnπ3+isinnπ3)sinnπ3=0nπ3=mπn=3m2018nmin=2019
Commented by gunawan last updated on 02/Mar/19
Nice thank you Sir
NicethankyouSir
Commented by malwaan last updated on 02/Mar/19
what is the point ?  we can write 2019  without proof   please explane sir
whatisthepoint?wecanwrite2019withoutproofpleaseexplanesir
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.
n=3m2018meansnmustbedivisibleby3andatleast2018.anumberisdivisibleby3ifthesumofitsdigitsisdivisibleby3.with2018thesumofitsdigitsis2+0+1+8=11,notdivisibleby3,but2+0+1+9=12,itsdivisibleby3,thereforenmin=2019.
Commented by malwaan last updated on 03/Mar/19
thank you very much sir
thankyouverymuchsir
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?
siriheardthisforcheckingdivisibilitywith3wehavetoaddthedigitsofthenumberspecialcase:ifadigitcontains9wecandropthe9andaddotherdigitseg:20192+0+1=3whichisdivisibleby3.whyisthissir?
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.
becausedigit9isalreadydivisiblyby3.ifthesumofotherdigitsisdivisibleby3,thenthetotalsumofalldigitsisalsodivisibleby3andthenumberisdivisibleby3.infactyoucandropnotonlydigit9,butalsodigits6,3and0,andonlyneedtoaddotherremainingdigits.e.g.wetakethenumber123450906497628005weonlyneedtoaddthebluedigits:1+2+4+5+4+7+2+8+5=38whichisnotdivisibleby3,thereforethegivennumberisnotdivisibleby3.
Commented by Kunal12588 last updated on 03/Mar/19
great sir. and thanks
greatsir.andthanks

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