a number is given with three bases base2,base 3,base5 such that last digit of the number in all the three cases is 1,and the leading digit is 1 in exactly 2 out of 3 cases.find the number?

• 91 is the right answr
  1)31(base2)=11111
  31(base3)=1011
  31(base5)=111
  2)91(base2)=1011011
  91(base3)=10101
  91(base)=331
  condition is that last is 1 all bases and leading digit(first digit) is 1 in exactly 2 out of 3 cases.
  31 has the leading digit 1 in all the cases .so 31 is wrong ( leading digit should be 1 in exactly 2 cases not more than that )
• 10 years agoHelpfull: Yes(31) No(3)
• divide 31 by 2 3 and 5 u will got both the condition satisfied.....and the another option is 91 divide by 5 you will got one 3.....so it is not satisfy the one condition........
• 10 years agoHelpfull: Yes(2) No(4)
• i think 31 is the answer
  
• 10 years agoHelpfull: Yes(1) No(3)