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sir-number-of-3-digit-numbers-which-are-divisible-by-a-3-b-4-c-6-d-7-e-8-f-9-g-11-when-repetetion-is-1-Allowwd-2-Not-allowed-kindly-help-me-sir-




Question Number 197564 by SLVR last updated on 21/Sep/23
sir...number of 3 digit  numbers which are divisible  by   a)3  b)4  c)6  d)7  e)8  f)9  g)11  when repetetion is  1)Allowwd  2)Not allowed..  kindly help me sir
sirnumberof3digitnumberswhicharedivisiblebya)3b)4c)6d)7e)8f)9g)11whenrepetetionis1)Allowwd2)Notallowed..kindlyhelpmesir
Commented by Tinku Tara last updated on 21/Sep/23
All numbers are divisble by one.  check your question
Allnumbersaredivisblebyone.checkyourquestion
Commented by SLVR last updated on 21/Sep/23
i am sorry by 1 was wrogly typed
iamsorryby1waswroglytyped
Commented by JDamian last updated on 22/Sep/23
  In the case 1),   [A] the 3−digit numbers span  100 to 999.  [B] the numbers which are divisible  by  n  appears every n positions.    With these hints I presume you are  smart enough to calculate the values  you asked for.
Inthecase1),[A]the3digitnumbersspan100to999.[B]thenumberswhicharedivisiblebynappearseverynpositions.WiththesehintsIpresumeyouaresmartenoughtocalculatethevaluesyouaskedfor.
Commented by mr W last updated on 24/Sep/23
2) repetition not allowed  a) divisible by 3       ⇒see example in Q197589    numbers of type pqr: 30×3!=180  numbers of type p0q: 2×(3+3×3)×2!=48  totally: 180+48=228 ✓
2)repetitionnotalloweda)divisibleby3seeexampleinQ197589numbersoftypepqr:30×3!=180numbersoftypep0q:2×(3+3×3)×2!=48totally:180+48=228
Answered by Rasheed.Sindhi last updated on 22/Sep/23
(1)Repetition allowed_     If x is 3-digit number then 100≤x≤999  If x is 3-digit number divisible by d then  ⌈((100)/d)⌉×d≤x≤⌊((999)/d)⌋×d and x is an AP  with common difference d  If a is first term a_n  is last term(nth term)  a_n =a+(n−1)d  ⌊((999)/d)⌋×d=⌈((100)/d)⌉×d+(n−1)d  ⌈((100)/d)⌉+(n−1)=⌊((999)/d)⌋  n=⌊((999)/d)⌋−⌈((100)/d)⌉+1  For example:  (a) When d=3  n=⌊((999)/3)⌋−⌈((100)/3)⌉+1=333−34+1=300  So there are 300   3-digit numbers which  are divisible by 3  (b) When d=4  n=⌊((999)/d)⌋−⌈((100)/d)⌉+1    =⌊((999)/4)⌋−⌈((100)/4)⌉+1=249−25+1=225  (c) When d=6  n=⌊((999)/d)⌋−⌈((100)/d)⌉+1    =⌊((999)/6)⌋−⌈((100)/6)⌉+1=166−17+1=150
(1)RepetitionallowedIfxis3digitnumberthen100x999Ifxis3digitnumberdivisiblebydthen100d×dx999d×dandxisanAPwithcommondifferencedIfaisfirsttermanislastterm(nthterm)an=a+(n1)d999d×d=100d×d+(n1)d100d+(n1)=999dn=999d100d+1Forexample:(a)Whend=3n=99931003+1=33334+1=300Sothereare3003digitnumberswhicharedivisibleby3(b)Whend=4n=999d100d+1=99941004+1=24925+1=225(c)Whend=6n=999d100d+1=99961006+1=16617+1=150

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