Solve:CAT=(C+A+T)*C*A*T

• a little correction:For T=5
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  100C+ 10A +T = (C+A+T)*C*A*T => 100+ 10(A/C) +(T/C)= (C+A+T)*C*A*T..Now clearly (A/C)100+10A+5 = (6+A)*A*5 => 21+2A=(6+A)*A => (21/A)+ 2 = 6+A => 21/A = A+4 => A=1,3,7 => A=3 ...
  So, C=1,A=3,T=5..
  
  135=(1+3+5)*1*3*5..
• 11 years agoHelpfull: Yes(8) No(0)
• C*C=C;A*A=A;T*T=T;
  other combination results 1
  therefore (C+A+T)*C*A*T= CCAT+CAAT+TTCA=C+A+T= CAT
• 11 years agoHelpfull: Yes(3) No(2)
• a little correction:
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  if T=5,then 20C +2A +1=(C+A+5)*A*C,=> C divides 2A+1,let, 2A+1=KC,so eqn can be re written as 40C +2KC= (2C+ KC +9)*A*C,=> 40+2K ={C(K+2)+9}*A,=> A
• 11 years agoHelpfull: Yes(2) No(0)
• Here clearly C,A,T represent 3 different digits from (0,1,2,....,9).
  CAT=100C+10A+T =(C+A+T)*C*A*T;So last_digit(l.d) of {(C+A+T)*C*A} is 1.
  so,l.d of (A*C) can be 1,3,7,9..
  if l.d of (A*C) =1, then possible values of (A,C)=(3,7),(7,3)
  then l.d of (C+A+T) must be 1 => T=1.so, (A,C,T)=(3,7,1),(7,3,1).
  but then CAT is nt = to (C+A+T)*C*A*T
  if l.d of (A*C)=9,then (A,C)=(1,9),(9,1).SO,L.D OF (C+A+T) must be 9 =>T=9 but A,C,T assume different values..So,this case is also not possible..
  if l.d of(A*C)=3,THEN(A*C)=(3,1),(1,3),(7,9)(9,7).so,l.d of (C+A+T) must be 7 so,we get T=3,when (A*C)=(3,1),(1,3)..we reject this case obviously.
  T=1,when (A*C)=(7,9)(9,7),for this case we see (C+A+T)*C*A*T=17*63=1071,so we also reject this case.
  Lastly l.d of (A*C)=7=> (A,C)=(1,7),(7,1),(3,9),(9,3),SO here l.d of (C+A+T) must be 3,=> T=5 for (A,C)=(1,7),(7,1)
  & T=1 for (A,C)=(3,9),(9,3)
  for T=5 ,(A,C)=(1,7),(7,1)=> LHS =13*35=455,so for this case also no solution..
  
  for T=1,(A,C)=(3,9),(9,3)=> LHS = 13*27=351,so no sln for this case also..
  
  again ,since 5*5=25, 6*6=36, so,
  if T=5 then l.d of (C+A+T)*C*A can be 5.=> l.d(C+A+T)=5=> (C+A)=10,& l.d of (C*A)=1 => (A,C)=(3,7),(7,3).But then LHS >1000.
  
  Lastly,T=6 =>6 divides 100C+10A+T=> 3 divides (A+C);SO (A+C)=3,6,9,12,15.
  17C+2A -(A+C)/3 +1 = (A+C+6)*C*A,since 3 divides (A+C+6) so, 3 divides {4(A+C)/3 -1}; => 9 divides [4(A+C)-3] =>(A+C)=3,12.
  SO, 6C+A -1 = 3CA ...=>5C+3-1=3CA => 5C+2 =3CA yields no solution..
  the other case 6C+A -5= 6CA, => 6 divides (A-5).=>A=5.Still it yields no soluiton..
  so,the system has no solution....
  
• 11 years agoHelpfull: Yes(1) No(0)
• CAT=(C+A+T)*C*A*T
  CAT=(C+A+T)CAT
  C+A+T=1
• 11 years agoHelpfull: Yes(1) No(1)
• C^2+2AC+2TC+2AT+T^2
• 11 years agoHelpfull: Yes(0) No(0)