Sum of a positive single digit number, its square & its cube is 3 more than a perfect square number. Sum of all such single digit numbers will be

Note: 0 is considered as a perfect square number

• sqrt(a+a^2+a^3-3)=a natural number
  1,3,4 are the numbers
  so Sum of the numbers=1+3+4=8
• 10 years agoHelpfull: Yes(2) No(0)
• answer is 8 as given by ram, here is my explanation
  according to question we can write
  1+1^2 + 1^3 -3 = 3-3 = 0, which is a perfect square
  2+4+8 -3 = 11,
  3+9+27 - 3 = 36
  4+16+64- 3 =81
  5+25+125 - 3 =152
  6+36+216 -3 = 255
  7+49+343 - 3 = 396
  8+64+512 -3 = 584
  9+81+729 -3 = 816
  
  it is clear from above that only 1,3,4 gives perfect square
  so the sum will be 1+3+4 = 8
• 10 years agoHelpfull: Yes(0) No(0)
• From the question, we can write
  1+1^2 + 1^3 -3 = 3-3 = 0, which is a perfect square
  Similarly the other possibilities are
  3 + 9 + 27 - 3 = 36
  and 4 + 16 + 64 - 3 = 81
  
  Hence the answer will be : (1+3+4) = 8(answer)
• 10 years agoHelpfull: Yes(0) No(0)