If x>y>125, sum of cube of that numbers ??
1) always divisible by x
2)always divisible by y
3)always divisible by 3
4)all of the above.

• Complete question is
  D=X^3-Y^3
  than D-1 is divisible by
  1) always divisible by x
  2)always divisible by y
  3)always divisible by 3
  4)all of the above.
  
  ANS=
  x^3-y^3=(x-y)^3+3*x^2*y-3*x*y^2;
  since x & y is consecutive whole no so x-y=1;
  D-1=3*x^2*y-3*x*y^2
  3*x*y*(x-y)
  
  So D-1 is always divisible by x,y and 3
  
  hence ans is D
• 10 years agoHelpfull: Yes(26) No(1)
• all have written the wrong ques...the ques was to see if the difference of the digits of the cube...so the ans is none of the above...
• 10 years agoHelpfull: Yes(3) No(1)
• 3. Always divisable by 3
• 10 years agoHelpfull: Yes(2) No(5)
• All the above.... Test for 1000,1001 or any number pair
• 10 years agoHelpfull: Yes(2) No(0)
• answer is d.(all of the above)
  7^3>6^3>5^3
  343-256=127
  127-1=126 which is divisible by all.
• 10 years agoHelpfull: Yes(1) No(2)
• Akash kumar D-1 = 3*x*Y - 1
  and D= 3XY
  
• 10 years agoHelpfull: Yes(1) No(1)
• D=X^3-Y^3= (X-Y)^3-3XY(X-Y)
  since X and Y are two consecutive numbers.
  so, X-Y=1
  
  D=1+3XY
  D-1=3XY
  hence D-1 divisible by X,Y and 3.
  
• 10 years agoHelpfull: Yes(0) No(0)