(logA + logB +logC)/log6=6; find how many possible solution exist.

• Rajnikanth the way u done is correct but
  log(A*B*C) = 6*log 6 then log(A*B*C) = log(6^6)
  
  so A*B*C = 6*6*6*6*6*6
  so no of possibilities are
  (6*6,6*6,6*6) = 1
  (6*6*6,6*6,6) = 6
  (6*6*6*6,6,6) = 3
  so total 10 ways i think
• 10 years agoHelpfull: Yes(32) No(2)
• (log A + log B + log C)/log 6 = 6
  =>(log (A*B*C))/log 6 = 6
  =>(log (A*B*C))=6*log 6
  =>(log (A*B*C))=log 36
  Therefore A*B*C=36
  
  So no.of possible solution will be 8
  1*1*36
  1*2*18
  1*3*12
  1*4*9
  1*6*6
  2*2*9
  2*3*6
  3*3*4
• 10 years agoHelpfull: Yes(10) No(36)
• (logA+logB+logC)/ log6 =6
  
  log(ABC)=6*log6
  ABC=6^6
  
  we can factorize 6 as (1*2*3) or(1*1*6)
  
  case 1: when solutions are different (at least two)
  possible combinations (1,2,3)= 6 solutions
  or (1,1,6)= 3 solutions
  
  case 2: when all solutions are equal i.e. (A*B*C)=(6*6, 6*6, 6*6)= 1 solution
  
  hence total no. of solutions= 6+3+1= 10
  
• 10 years agoHelpfull: Yes(10) No(2)
• 6*log6=36.....????
  
• 10 years agoHelpfull: Yes(3) No(3)
• Ans will be 14! / (6! 6! 2!)
  Explanation:
  
  ABC = 2^6 * 3^6
  The question now boils down to dividing 6 no. of 2's and 6 no. of 3's among A, B & C.
  
  The arrangement is equivalent to possible arrangement of followings -
  
  0 0 0 0 | 0 0 * * | * * * * (this is similar to integral solution question i.e. A + B + C = 12 etc.)
  
  No. of possible arrangement is 14! / (6! 6! 2!)
• 9 years agoHelpfull: Yes(2) No(1)
• Rajnikant bhai.sachmuch rajnikant ban gaya..6log6= 36....how is it possible rajnikant jee??
• 9 years agoHelpfull: Yes(1) No(1)
• I have a query:
  Why we considering only combinations in powers of 6
  Can we write A*B*C as 2^6 * 3^6
  So we can also have A = 1, B = 2 , C= 2^5*3^6?????
  And so we can form different combinations with distributed powers of 2 and 3.
• 9 years agoHelpfull: Yes(1) No(3)
• A*B*C = 12
  A,B,C are different numbers so > 1*4*3 = 12 , 2*6*1 = 12, and both can be arranged by 3 ways so ans must be 6.
• 10 years agoHelpfull: Yes(0) No(7)
• log(A*B*C) = 6*log 6 then log(A*B*C) = log(6^6)
  
  so A*B*C = 6*6*6*6*6*6
  
  no of ways of dividing 6 sixes among 3 persons a,b,c is n+r-1 C r-1
  
  6+3-1 C 3-1==>8C2==>28
  
  so 28 solutions are possible
  
• 8 years agoHelpfull: Yes(0) No(1)
• log(A*B*C)=6*log(6)
  =log(6^6)
  =log(6*6*6*6*6*6)
  A*B*C=(2^6)*(3^6)
  if a^n * b^m then no of solution is (n+1)*(m+1)
  so here ans is 7*7=49
  no of solution =49
• 8 years agoHelpfull: Yes(0) No(0)