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Question-a-which-numbers-have-an-odd-number-of-divisors-b-Is-there-a-number-with-exactly-13-divisors-c-Generalize-




Question Number 164300 by otchereabdullai@gmail.com last updated on 16/Jan/22
Question  a.which numbers have an odd number       of divisors  b.  Is there a number with exactly 13        divisors  c.  Generalize
Questiona.whichnumbershaveanoddnumberofdivisorsb.Isthereanumberwithexactly13divisorsc.Generalize
Answered by mr W last updated on 16/Jan/22
a.  all numbers have odd number of  divisors, if in their prime   factorization all exponents are  even. i.e. n=p_1 ^(2k_1 ) p_2 ^(2k_2 ) p_3 ^(2k_3 ) ... with  p_i ∈P, k_i ∈N.  this can be easily checked. the number  of divisors of n is  N=(2k_1 +1)(2k_2 +1)(2k_3 +1)...  which is always an odd number.    b.  yes!  e.g. 2^(12) , 3^(12) , 11^(12)  etc. have exactly 13  divisors.  with knowledge from above, we only  need to find n such that  N=(2k_1 +1)(2k_2 +1)(2k_3 +1)...=13  since 13 is prime, therefore the only  possibility is k_1 =6, k_(≥2) =0. that means  all numbers in form n=p^(12)  has exactly  13 divisors, where p is any prime  number.    c.  n=p^(12)  with p∈P
a.allnumbershaveoddnumberofdivisors,ifintheirprimefactorizationallexponentsareeven.i.e.n=p12k1p22k2p32k3withpiP,kiN.thiscanbeeasilychecked.thenumberofdivisorsofnisN=(2k1+1)(2k2+1)(2k3+1)whichisalwaysanoddnumber.b.yes!e.g.212,312,1112etc.haveexactly13divisors.withknowledgefromabove,weonlyneedtofindnsuchthatN=(2k1+1)(2k2+1)(2k3+1)=13since13isprime,thereforetheonlypossibilityisk1=6,k2=0.thatmeansallnumbersinformn=p12hasexactly13divisors,wherepisanyprimenumber.c.n=p12withpP
Commented by Rasheed.Sindhi last updated on 16/Jan/22
Nice & Elegant!
Nice&Elegant!
Commented by otchereabdullai@gmail.com last updated on 16/Jan/22
God bless you prof W
GodblessyouprofW
Commented by Tawa11 last updated on 16/Jan/22
Great sir
Greatsir
Commented by nikif99 last updated on 16/Jan/22
Commenting (a) I would say that all  perfect squares have odd number  of divisors.  ∵ Divisors of N are in pairs (p, N/p).  If N=perfect square, (√N) has no pair.
Commenting(a)Iwouldsaythatallperfectsquareshaveoddnumberofdivisors.DivisorsofNareinpairs(p,N/p).IfN=perfectsquare,Nhasnopair.

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