When numbers are written in base b, we have 12 x 25 = 333, the value of b is?
a) 8
b) 6
c) None
d) 7

• ans d) 7
  12 ,25 and 333 in decimal form can be written as (1*b^1 + 2*b^0) , (2*b^1 + 5*b^0), (3*b^2+3*b^1 + 3*b^0)
  
  12 *25 = 333
  (1*b^1 + 2) * (2*b^1 + 5*) = (3*b^2+3*b^1 + 3)
  b^2+ 9b + 10 = 3b^2 + 3b + 3
  b^2 -6b -7 = 0
  b^2 - 7b + b - 7 =0
  (b-7)(b+1)=0
  b=7 or b=-1
  base cannot be a negative value so b=7
• 9 years agoHelpfull: Yes(20) No(3)
• option d 7.
• 9 years agoHelpfull: Yes(3) No(1)
• (1*b^1+2)*(2*b^1+5)=(3*b^2+3*b+3)
  
  (b+2)(2b+5)=3b^2+3b+3
  
  b^2 -6b -7=0
  b=7 or b=-1
• 9 years agoHelpfull: Yes(3) No(0)
• ans d) 7
  12 ,25 and 333 in decimal form can be written as (1*b^1 + 2*b^0) , (2*b^1 + 5*b^0), (3*b^2+3*b^1 + 3*b^0)
  
  12 *25 = 333
  (1*b^1 + 2) * (2*b^1 + 5*) = (3*b^2+3*b^1 + 3)
  b^2+ 9b + 10 = 3b^2 + 3b + 3
  b^2 -6b -7 = 0
  b^2 - 7b + b - 7 =0
  (b-7)(b+1)=0
  b=7&-1
  
• 8 years agoHelpfull: Yes(1) No(2)
• answer is 7
• 8 years agoHelpfull: Yes(0) No(0)
• Ans is 7 [(1*b^1+2)(2*b^1+5)=(3*b^2+3b^1+3)]
• 8 years agoHelpfull: Yes(0) No(0)
• 7
  
  
  (b+2)(2b+5)=3b^2+3b+3
• 8 years agoHelpfull: Yes(0) No(0)