When the number is written in base b, we have 12*25 =333. The
value of b is ?

• 12*25=333..
  (1*b^1+2*b^0)*(2*b^1+5*b^0)=
  (3b^2+3b^1+3b^0)
  b=7....
• 9 years agoHelpfull: Yes(18) No(2)
• ans-7
  12*25=333
  (1*b^1+2*b^0)* (2*b^1+5*b^0)=(3*b^2+3*b^2+3*b^0)
  (b+2)(2b+5)=3b^2+3b+3
  2b^2+5b+4b+13b^2+3b+3
  b^2-6b-7=0
  b=7,-1
  base can't be negative so ans is 7
• 9 years agoHelpfull: Yes(11) No(0)
• Ans is:7
  Take option to solve it is best way..
  Condition is LHS==RHS
  12*25..
  1*7+2=9
  2*7+5=19
  9*19=171
  Now 333
  3*7^2+7*3+3=171
  So LHS==RHS
• 9 years agoHelpfull: Yes(6) No(1)
• convert all the num to base x
  (1*x+2*x^0)*(2*x+5*x^0)=(3*x^0+3*x^1+3*x^2)
  solving X=7
• 9 years agoHelpfull: Yes(2) No(0)
• R u sure this is the actual question?
  It's coming out -1 and base can't be in negative. I think smthng is missing
• 9 years agoHelpfull: Yes(1) No(3)
• Let the base = b
  So, (b+2)(2b+5) = (b+2)(2b+5)=3b2+3b+3
  2b2+9b+10=3b2+3b+3
  b2−6b−7=0
  Solving we get b = 7 or -1
  So b = 7
  
• 9 years agoHelpfull: Yes(1) No(0)
• 12*25=333
  (1*b^1+2*b^0)*(2*b^1+5*b^0)=(3*b^2+3*b^1+3*b^0)
  (b+2)*(2b+5)=(3b^2+3b+3)
  b^2-6b-7=0
  b=7
• 9 years agoHelpfull: Yes(0) No(0)
• value of b is 7
  (b+2)*(2b+5)=3b^2+3b+3
  by solving the above equation we get ans as 7
• 9 years agoHelpfull: Yes(0) No(0)