A number p is 3 digit number like 1st digit is square of last digit. and no digit is same . so how many p is possible which is divisible by 3 .

• Three digit number = xyz
  Rule1: 1st digit is square of last = x^2yx
  
  Only numbers and their square are in single digit= 2,3 and their square are 4&9.
  
  possible numbers:
  4y2=402,412,432,452,462,472,482,492
  9y3=903,913,923,943,953,963,973,983
  
  Rule2: Number should be divisible by 3.
  Now all possible numbers are:
  4y2/3=402,432,462,492 = 4 count
  9y3/3=903,963 = 2 count
  
  hence total number count is 4+2=6.
• 9 years agoHelpfull: Yes(85) No(3)
• total 6
  903
  963
  402
  432
  462
  492
  
• 9 years agoHelpfull: Yes(24) No(0)
• i dont think unit digit squares are only 2 & 3 even 1 have.(0,1,2,3,4,5,6,7,8,9)
  1^2=111,141,171 which is 3 digit div by 3
  2^2=402,432,462,492 which is div by 3
  3^2=903,933,963,993
  so total=11
• 9 years agoHelpfull: Yes(10) No(27)
• This is not for E-litmus level question.
• 9 years agoHelpfull: Yes(5) No(4)
• The three digit number is of the form a^2ba.
  Now the numbers in the units digit can be (1,2,3) and in tens place can be (1,4,9) respectively.
  Now when in units place 1 is there we get: 111, 141, 171 which are divisible by 3.
  Now when in units place 2 is there we get: 402, 432, 462, 492 which are all divisible by 3.
  Now when in units place 3 is there we get: 903, 933, 963, 993 which are again all divisible by 3.
  
  Hence total of 11 numbers..
• 9 years agoHelpfull: Yes(4) No(5)
• ans. 6
  913
  923
  943
  953
  973
  983
• 9 years agoHelpfull: Yes(2) No(9)
• only 2 posible no. can take 1st position ie- 2 and 3with last digit being 4&9.
  
  now to find the middle no combination, we know that for a no. divisible by 3.. sum of digit shud be divisible by 3.
  So, 2+6 & 3+9 are already divisible by 3.
  Now middle digit can be 0,3,6,9...
  ie, no are 402,432,462,492,903,963.
  
  so ans is 6
• 9 years agoHelpfull: Yes(2) No(0)
• total 8
  903
  933
  993
  963
  402
  432
  462
  492
• 9 years agoHelpfull: Yes(2) No(18)
• there are 3 possibilities for last digit.....
  which is 1,2,3..
  for digit 1.. 111,141,171...total 3 numbers..
  for digit 2....402,432,462,492...total 4 numbers..
  for digit 3..903,933,963,993...total 4 numbers..
  answer is =3+4+4=11
  
• 9 years agoHelpfull: Yes(2) No(2)
• ans . 6
  903
  923
  432
  462
  492
  936
  943
• 9 years agoHelpfull: Yes(1) No(6)
• only 6 no is possible
  4m2,9n3
  this is general no format sum of these no 4+m+2 will be divisible by 3
  similarly others no
  so only 402,432,462,492
  903,963
• 9 years agoHelpfull: Yes(1) No(0)
• answer is 6
  
• 9 years agoHelpfull: Yes(1) No(0)
• The answer will be 11.
  condition 1: 1st digit is square of last digit.
  The possible 1st and last digit combination will be : 1st-1, last-1
  1st-4, last-2
  1st-9, last-3
  condition 2: The number should divisible by 3.
  Therefore, for 1st=1 and last=1, the possible number will be- 111, 141, 171.
  
  for 1st=4 and last=2, the possible number will be- 402, 432, 462, 492.
  
  for 1st=4 and last=2, the possible number will be- 903, 933, 963, 993.
  Hence total number of combination will be: 3+4+4=9.
  
• 9 years agoHelpfull: Yes(1) No(5)
• Ans 5
  432
  462
  492
  903
  963
• 9 years agoHelpfull: Yes(0) No(5)
• 804,834,864,894 can also satisfy
• 9 years agoHelpfull: Yes(0) No(11)
• ans is 6
  
  4_2
  9_3
  are the possibilities.....( 402,432,462,492
  903,963 )
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 6
  903, 963,402,432,462,492
• 9 years agoHelpfull: Yes(0) No(0)
• The correct answer is 6
  let the number P=xyz
  1st it is said that=z^2yz
  Only numbers and their square are in single digit =1,2,3 and their square are 1,4,9
  we will not take 1 as in question it is said that no digit is same.
  
  Possible value:
  2^2y2=402,,432,462,492 =4
  
  3^2y2 =903,963=2
  so total numbers are 6.
  
• 9 years agoHelpfull: Yes(0) No(0)
• Total is 8....
  402,432,462,492
  903,933,963,993
• 9 years agoHelpfull: Yes(0) No(1)
• 11
  111,141,171
  402,462,432,492
  90,933,963,993
• 9 years agoHelpfull: Yes(0) No(0)
• 3 ways for no 4 _ 2 and 4 ways for 9 _ 3 which is divisible by 3. so,
  total of 3+4 = 7 ways and 7 possible no is there with non repeating three digit no.
• 9 years agoHelpfull: Yes(0) No(0)
• 6
  402
  432
  462
  492
  903
  963
• 9 years agoHelpfull: Yes(0) No(0)
• possible numbers are:- 402,432,462,492,903,963,
  so total no can be possible only :- 6.
• 9 years agoHelpfull: Yes(0) No(0)
• Rashmi or minupala boki h...
• 9 years agoHelpfull: Yes(0) No(0)
• 402
  432
  462
  492
  903
  963
• 9 years agoHelpfull: Yes(0) No(0)
• 903
  963
  402
  432
  462
  492
  Total no of p=6
• 9 years agoHelpfull: Yes(0) No(0)
• No digit should be same..
  4_2 :- 402,432,462,492..total 4
  9_3 :- 903,963.. total 2
  total number is 6
• 9 years agoHelpfull: Yes(0) No(0)
• right answer 8 number
  402
  432
  462
  493
  903
  933
  963
  993
  and 111 not including because in question say no digit are same
• 9 years agoHelpfull: Yes(0) No(0)
• 1^1=1(rejected.since the numbers in the digits cannot repeat itself according to the question)
  2^2=4
  3^2=9
  so the 3 digit numbers will be- 4y2 & 9y3
  check y=0,1,2....9 and check the numbers divisibility by 3 such that every digit in the 3 digit number is unique
  the numbers divisible by 3 are: 402,432,462,492,903,963 are the required numbers.
  
• 9 years agoHelpfull: Yes(0) No(0)
• 402,432,462,492,903,963=6 numbers
• 8 years agoHelpfull: Yes(0) No(0)