When numbers are written in base b, we have 12 * 25 = 333. The value of b is
(a)8
(b) 6
(c) None of these
(d) 7

• when a no. is written in base b
  (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)
  solve it further a quadratic eq. is formed
  
  b^2-6b-7
  after solving
  (b-7)(b-1)
  therefore b=7
  b=7
• 11 years agoHelpfull: Yes(23) No(2)
• just multiply last digits of both numbers, it should be last digit of product i.e. 3.
  2*5=3
  so 10=3
  which is only possible for base 7.
• 11 years agoHelpfull: Yes(10) No(1)
• can u xpln it...pls
  
• 11 years agoHelpfull: Yes(1) No(1)
• b=7 is the answer
  
• 11 years agoHelpfull: Yes(0) No(1)
• (b^1+2)*(2b^1+5)=(3b^2+3b^1+3)
  2b^2+9b+10=3b^2+3b+3
  b^2-6b-7=0
  (b-7)(b+1)=0
  b=7
  ans is option d) 7
• 11 years agoHelpfull: Yes(0) No(1)
• (1*b ^1)+(2*b^0)*(2*b^1+5*b^0)=3*b^2+3*b^1+3*b^0
  ans is 7
• 11 years agoHelpfull: Yes(0) No(2)
• answer is 7.
  
• 11 years agoHelpfull: Yes(0) No(1)