a two digit num "ab" is six times the sum of its digits.find the value of b^a.

• two digit number = (10*a + b)
  sum of its digits = a + b
  then,
  (10*a + b) = 6*( a + b)
  10a - 6a = 6b - b
  a/b = 5/4
  a=5 and b=4
  the no is 54, which is 6 * (5 + 4).
  b^a = 4^5=4x4x4x4x4
  = 1024
  
  
• 13 years agoHelpfull: Yes(32) No(1)
• according to question-> ab=6(a+b)
  multiples of 6 are 6,12,18,......,42,48,54,60.
  where 5+4=9->9*6=54 which gives the sol for question.
  here a=5 b=4 then b^a=4^5=1024 (ans)
• 13 years agoHelpfull: Yes(9) No(0)
• Two digit number will be 10a+b which is equal to 6(a+b).(As per given).
  i.e 10a+b=6(a+b)
  on solving we well get 4a=5b.
  this will be equal only for a=5 and b=4.
  so 4^5=1024.
  1024 is answer.
• 13 years agoHelpfull: Yes(2) No(0)
• two dizit no = 10a+ b
  given that 10a +b = 6(a+b)
  which means 4a = 5b
  a/b = 5/4
  a: b = 5: 4
  so number is 54
  now b^a=4^5 =1024
• 13 years agoHelpfull: Yes(2) No(0)
• 20ans.[eqtn{(10a+b)=6(a+b)} = a/b=5/4 =>5*4=20ans.]
• 13 years agoHelpfull: Yes(1) No(13)
• 4^5=1024
  ab=6(a+b)
  54=6(5+4)
  54=54
• 13 years agoHelpfull: Yes(0) No(4)