There is a 3 digited number. 3rd number is the square root of the 1st digit.
2nd digit is the sum of 1st and 3rd.And that number is divisible by 2,3,6,7.
What is that number?

• Assume that the digit is taken as XYZ.
  Condition 1=> Z=root(x). Possible digits of x is 1,4,9
  2=> Y=x+z
  X Y Z
  1 2 1
  4 6 2
  9 12 3 wrong because it is 4 digit.
  
  and 121 is not divisible by 2
  
  The answer is 462.
  
• 12 years agoHelpfull: Yes(18) No(2)
• 462 is desired number.
• 12 years agoHelpfull: Yes(9) No(2)
• let 3digit number be XYZ... so acc to prblm Z=sqrt(X),Y=X+Z.let X=4 so Z=2 and Y=6 hence 3digit will be 462 which is divisible by 2,3,6,7
• 12 years agoHelpfull: Yes(3) No(0)
• 462.
  Third number is the square root of first digit.. square root of 4 is 2.
  sum of first and third digit is 4+2=6
  then 462 is divisible by 2,3,6 and 7
• 12 years agoHelpfull: Yes(3) No(0)
• 
  l.c.m of 2,3,6,7 is 42.
  3rd digit 2 whch is sq rt of 1st digit 4 and 2nd digit is 6 whch is sum of 2+4.
  so the 3 digit num is 462 whch is divisible by 42.
• 12 years agoHelpfull: Yes(3) No(0)
• 462 divisible by 2,3,6,7
• 11 years agoHelpfull: Yes(0) No(0)
• first, 3 digit num is divisible by 2,3,6,7 so unit place must b even num.
  from this 2*2=4,3*3=9,4*4=16
  sqrt of 4 is not apt nd 3 too.
  so unit place is 2.
  sqrt of 2 is 4
  ans is 462
• 11 years agoHelpfull: Yes(0) No(0)
• ok the no must be an lcm of all the divisible nos so lcm of 2,3,6,7 is 252
• 10 years agoHelpfull: Yes(0) No(3)