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Given-function-f-4567-321567-567-321-888-f-32156-12062-156-120-276-find-the-value-of-f-20-22-21-




Question Number 198447 by cortano12 last updated on 20/Oct/23
  Given function     f(4567,321567)= 567+321=888.    f(32156,12062)= 156+120=276    find the value of        f(((20^(22) )/(21)) ).
Givenfunctionf(4567,321567)=567+321=888.f(32156,12062)=156+120=276findthevalueoff(202221).
Commented by mr W last updated on 20/Oct/23
the given function is f(x,y), but you   are asking f(a)=?  it makes no sense to me.
thegivenfunctionisf(x,y),butyouareaskingf(a)=?itmakesnosensetome.
Commented by Rasheed.Sindhi last updated on 20/Oct/23
sir, I think comma is   decimal separater here.
sir,Ithinkcommaisdecimalseparaterhere.
Commented by cortano12 last updated on 20/Oct/23
the questions mean if    f(((157683)/(321)))=f(491,785047)=491+785
thequestionsmeaniff(157683321)=f(491,785047)=491+785
Commented by mr W last updated on 20/Oct/23
ok. “,”=“.”
ok.,=.
Commented by Rasheed.Sindhi last updated on 20/Oct/23
 f(((157683)/(321)))=f(491,785047)=491+785    But ((157683)/(321))=491.224299≠491,785047
f(157683321)=f(491,785047)=491+785But157683321=491.224299491,785047
Answered by Rasheed.Sindhi last updated on 20/Oct/23
  f(4567,321567)= 567+321=888    f(32156,12062)= 156+120=276       f(((20^(22) )/(21)) )=?  f(((20^(22) )/(21)) )=⌊((20^(22) )/(21))⌋mod 1000                    +⌊((20^(22) )/(21))×1000⌋mod 1000        ∵      ((20^(22) )/(21)) = (2^(22) /(21))×10^(22)   =⌊(2^(22) /(21))×10^(22) ⌋mod10^3 +⌊(2^(22) /(21))×10^(25) ⌋mod10^3              =619+47=666  Continue
f(4567,321567)=567+321=888f(32156,12062)=156+120=276f(202221)=?f(202221)=202221mod1000+202221×1000mod1000202221=22221×1022=22221×1022mod103+22221×1025mod103=619+47=666Continue
Answered by MM42 last updated on 20/Oct/23
20^(22)  ≡^(21) 1⇒20^(22) =21k+1 ;  k=abc...619  ⇒((20^(22) )/(21))=((21×abc..619+1)/(21))=abc..619+0.476..  ⇒f(((20^(22) )/(21)))=619+47=666 ✓
20222112022=21k+1;k=abc619202221=21×abc..619+121=abc..619+0.476..f(202221)=619+47=666
Commented by cortano12 last updated on 20/Oct/23
how to get k=abc...619?
howtogetk=abc619?
Commented by MM42 last updated on 20/Oct/23
20^(22) −1=a_1 ..a_n 9...999=21k  if  k=abc...xyz ⇒ 21×...xyz=...999  ⇒z=9⇒21×...xy9=...999     y+8=9⇒y=1  ⇒21×...x19=...999  x+3=9⇒x=6  ⇒k=abc...619  ✓
20221=a1..an9999=21kifk=abcxyz21×xyz=999z=921×xy9=999y+8=9y=121×x19=999x+3=9x=6k=abc619

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