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Let-abc-bca-cab-defg-where-a-b-g-are-decimal-digits-may-be-equal-to-0-Show-that-i-dg-a-b-c-ii-e-f-d-g-




Question Number 200836 by Rasheed.Sindhi last updated on 24/Nov/23
  Let abc ^(−) + bca ^(−) + cab ^(−) = defg ^(−)   where a,b,...,g are decimal digits  (may be equal to 0)   Show that  (i) dg ^(−) =a+b+c  (ii) e=f=d+g
Letabc+bca+cab=defgwherea,b,,garedecimaldigits(maybeequalto0)Showthat(i)dg=a+b+c(ii)e=f=d+g
Commented by LaChicaFrancesa last updated on 24/Nov/23
Hola, soy nueva en la app y quería consultar cómo se hace para que las 3 letras lleven encima la barra. Qué símbolo se usa?
Commented by AST last updated on 24/Nov/23
Do you mean abc^(−) +bca^(−) +cab^(−) ?
Doyoumeanabc+bca+cab?
Commented by Rasheed.Sindhi last updated on 24/Nov/23
Yes sir, exactly!
Yessir,exactly!
Commented by AST last updated on 24/Nov/23
dg^(−) =a+b+c is not true for a+b+c=19  since dg^(−) =29,For example a=9,b=8,c=2
dg=a+b+cisnottruefora+b+c=19sincedg=29,Forexamplea=9,b=8,c=2
Commented by Rasheed.Sindhi last updated on 25/Nov/23
(i)Use forum keyboard  (ii)Select the ′word′ which you  want make under  bar.  (iii) Tap on ■^(□)   (iv)Click on ′ − ′ (minus sign)
(i)Useforumkeyboard(ii)Selectthewordwhichyouwantmakeunderbar.(iii)Tapon◼◻(iv)Clickon(minussign)
Answered by AST last updated on 24/Nov/23
p=abc^(−) +bca^(−) +cab^(−) =111(a+b+c)≤2997⇒d≤2  defg^(−) =1000d+100e+10f+g≡g≡^(10) a+b+c  d=0⇒p≤999⇒a+b+c≤9  ⇒a+b+c=g=10(0)+g=10d+g=dg^(−)   d=1⇒10≤a+b+c≤18⇒a+b+c=10(1)+q≡^(10) q≡^(10) g  1≤q≤8⇒q=g⇒a+b+c=10(1)+g=10d+g=dg^(−)   d=2⇒19≤a+b+c≤27,   For 20≤a+b+c≤27;similar argument applies  Suppose a+b+c=19⇒defg^(−) =2109 ⇒dg^(−) =29≠19  This also contradicts (ii).
p=abc+bca+cab=111(a+b+c)2997d2defg=1000d+100e+10f+gg10a+b+cd=0p999a+b+c9a+b+c=g=10(0)+g=10d+g=dgd=110a+b+c18a+b+c=10(1)+q10q10g1q8q=ga+b+c=10(1)+g=10d+g=dgd=219a+b+c27,For20a+b+c27;similarargumentappliesSupposea+b+c=19defg=2109dg=2919Thisalsocontradicts(ii).
Commented by Rasheed.Sindhi last updated on 24/Nov/23
V. nice analysis!  Thanks sir!  The statement is not true in a   few cases, of course.However  your counter example may be   justified if we be allowed decimal  digits greater than 9 also.  Let X is digit having value10  982+829+298=1XX9     =1000+100(10)+10(10)+9=2109  Anyway you′re very right because  decimal digit may not exceed   than 9.
V.niceanalysis!Thankssir!Thestatementisnottrueinafewcases,ofcourse.Howeveryourcounterexamplemaybejustifiedifwebealloweddecimaldigitsgreaterthan9also.LetXisdigithavingvalue10982+829+298=1XX9=1000+100(10)+10(10)+9=2109Anywayyoureveryrightbecausedecimaldigitmaynotexceedthan9.

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