'P' is a 3 digit number. First digit from the left is a square of the last digit. None of the digit is repeated in 'P'. How many number of Primesubdents are possible. (Primesubdent is a number that can not be divided by 2,3,5)
a)6
b)
c)
d)72

• a) 6
  If P=xyz, given x=z^2 Also x,y & z are different
  x can be 4 & 9 i.e. square of 2 & 3 respectively,So P=4y2 & 9y3
  As P shouldn't be divisible by 2,3,5 we have only P=9y3
  So Combinations (913),(923),(943),(953),(973),(983) Total 6 are possible
• 11 years agoHelpfull: Yes(43) No(0)
• let P=xyz where x=z^2
  (z,x) may be (1,1),(2,4),(3,9)
  P will be of form 1y1,4y2,9y3
  given that digits are not repeated so 1y1 is not possible
  4y2 is also not possible bcoz its divisible by 2 so P is not Primesubdent
  
  so P=9y3 y=0,1,2,...,9
  P shouldn't be divisible by 2,3,5
  so P=(913),(923),(943),(953),(973),(983)
  
  total six numbers
  a)6
• 11 years agoHelpfull: Yes(32) No(0)
• From question last digit of 'P' is 2 when first digit is 4 or last digit of 'P'
  
  is 3 when first digit is 9.
  
  when last digit is 2 always divided by 2.
  
  when last digit is 3 is not divided by 3 for 913,923,943,953,973,983.
  
  so,number of primesubdents is 6.
• 11 years agoHelpfull: Yes(2) No(1)
• number is not divisible by 2,3,5 so last digits in p are 1,3,7 and first digit is sqr of last it can be only 3.
  the middle no can be chosen
  in 10 ways the answers is 10
• 11 years agoHelpfull: Yes(1) No(5)