Q. Find the 3 digit no. whose last digit is the squareroot of the first digit and second digit is the sum of the other two digits

• let xyz be the digits of no.,then
  z=sqrt(x)
  y=x+z=x+sqrt(x)
  so no. is =100x+10y+z
  =100x+10(x+sqrt(x))+sqrt(x)
  =110x+11*sqrt(x)
  if x=1, no. is =121
  if x=4, no. is =462
  if x=9, no. is =990+33=1023(not 3 digit)
  two no's 121 & 462
• 11 years agoHelpfull: Yes(41) No(0)
• 3 digit number
  so __ __ __
  in that gap we get
  (x) (x+sqrt(x)) sqrt(x)
  digits are 0,1,2,3,4,5,6,7,8,9
  sqrt applicable for 1,4,9
  first take 1 then number is 121
  second take 4 then number is 462
  third take 9 then number is 9123 which is not 3 digit numb
  so ans is 121 and 462
• 11 years agoHelpfull: Yes(11) No(2)
• let first digit be a then the number will be a (a+root(a)) root(a)
  if a=4
  then root(a)= 2
  a+root(a)=6
  for any other value of a value of a+root(a) will not be a single digit and the number has to be a 3 digit number
  so, the number will be 462
• 11 years agoHelpfull: Yes(5) No(1)
• single digit that must be square of other digit ,only 4 and 9 is on focus.
  hence as per given terms,that nuber looks somewhat
  4_2 or 9_3;
  but middle is sum of other two number hence,
  9+3=12 is not possible,
  the digit is
  462
• 11 years agoHelpfull: Yes(3) No(0)
• lets 3 digit no is xyz
  z=squareroot(x)
  y=x+z
  lets consider x=1
  z=squareroot(1)=1
  y=1+1
  so ans is 121
• 11 years agoHelpfull: Yes(1) No(0)
• digits are : x^2 x^2+x x
  possible no.: 1 2 1
  4 6 2
  
  ans is 121 & 462
• 11 years agoHelpfull: Yes(1) No(0)
• 462
  Last digit is square root of first digit
  And second digit is the sum of first and last
• 11 years agoHelpfull: Yes(0) No(0)
• let first term be x and last term be squareroot of x then middle term is (x)+ root(x). Then that 3 digit no must be 121 or 462
• 11 years agoHelpfull: Yes(0) No(0)
• 462 and 121
• 11 years agoHelpfull: Yes(0) No(0)
• either 121 or 462
• 11 years agoHelpfull: Yes(0) No(0)
• 462
  last digit=square root 4
  second digit=4+2=6
  so,ans 462
• 11 years agoHelpfull: Yes(0) No(0)
• Its a 3 digit no. So for a square root to be single digit we can consider 3 no.'s 1,4,9 having sq. root 1, 2, 3 respectively.. 9 should not be considererd bcoz sum of first & last digit(9+3=11) will not be single digit.
  We are left with 1 & 4.
  For 1, no. would be 121(1st digit, sum of 1st & last digit, squareroot of 1st digit)
  For 4, it would be 462 similarly
  Answer should be either 121 or 462
• 11 years agoHelpfull: Yes(0) No(0)
• As per question third digit may be 1,2,3.
  So first digit will be
  if last digit is 1-> 1_1,
  if last digit is 2-> 4_2,
  if last digit is 3-> 9_3.
  Given 2nd digit is sum of first and second, Therefore add first and last digit. With this condition last digit as 3 is not satisfied because 9+3 = 12. And condition for last digit 1, 2 satisfies wrt question.
  Ans- 121 (last digit sqr(1), second digit = sum of 1st and last i.e, 1+1 = 2)
  462 (last digit sqr(4), second digit = sum of 1st and last i.e, 4+2 = 6)
• 11 years agoHelpfull: Yes(0) No(0)
• answer will be 462
• 11 years agoHelpfull: Yes(0) No(0)
• 001, square root of 0+1 is 1 and sum of 0+1 is 1
• 10 years agoHelpfull: Yes(0) No(1)
• its 121 & 264
• 10 years agoHelpfull: Yes(0) No(0)