when number are written in base b,we have 12*25=333, the value of b is

• let base be x
  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
• 12 years agoHelpfull: Yes(20) No(2)
• 12 to the base b = 1*b¹+2*b⁰ = b+2 to the base 10
  25 to the base b = = 2b+5 " "
  
  12* 25 = (b+2)(2b+5) = 2b²+9b+10
  
  333 to the base b = 3b² +3b+3 to the base 10
  
  therefore 2b² + 9b+10 = 3b² +3b+3
  
  b² - 6b -7 =0
  b= 7,-1
  
  b=7
• 12 years agoHelpfull: Yes(20) No(3)
• please give the solution properly. How did you find the base to be 7 at the first place?
• 12 years agoHelpfull: Yes(9) No(3)
• (1*x+2*x^0)*(2*x+5*x^0)=(3*x^2+3*x^1+3*x^0)
  2x^2+9x+10=3x^2+3x+3
  x^2-6x-7=0
  solving X=7
• 12 years agoHelpfull: Yes(6) No(2)
• base 7.
  
  12*25 base 7 means 9*19 decimal equal to 171 which is 333 in base 7.
• 12 years agoHelpfull: Yes(3) No(1)
• plz explain clearly..
• 12 years agoHelpfull: Yes(1) No(5)
• let us assume base=b
  (b*1+2)(b*2+5)=3b^2+3b+3
  (b+2)(2b+5)=3b^2+3b
  b^2+9b+10=3b^2+3b+3
  b^2-6b-7=0
  b=7/b=-1
  ans:
  b=7
  b^
• 10 years agoHelpfull: Yes(1) No(0)
• base refers to 10. for eg: 12 can be written as 10+2 which is equal to b+2
  (b+2)(2b+5)=3b^2+3b+3
  2b^2+4b+5b+10=3b^2+3b+3
  b^2-6b-7=0
  solving the above eqn
  b=7 or b=-1
  the value of b cannot be negative and hence the value of b is 7
• 9 years agoHelpfull: Yes(1) No(0)
• please explain clearly..what method u used here..?
  
• 10 years agoHelpfull: Yes(0) No(1)
• Let the base = b
  So, (b+2) (2b+5) = (b+2)(2b+5)=3b2+3b+3(b+2)(2b+5)=3b2+3b+3
  2b2+9b+10=3b2+3b+32b2+9b+10=3b2+3b+3
  b2−6b−7=0b2−6b−7=0
  Solving we get b = 7 or -1
  So ANSWER will be the b = 7
• 7 years agoHelpfull: Yes(0) No(0)