Latest Elitmus Aptitude Question SOLUTION: 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.

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(#M40039527) ELITMUS Numerical Ability question Number System Keep an EYE

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.

Asked In Elitmus Prashant Baghel (9 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

AKASH KUMAR  (9 years ago)

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

Is this solution Helpfull? Yes (26) | No (1)  

abhishek maini  (9 years ago)

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...

Is this solution Helpfull? Yes (3) | No (1)  

Narasapu Hari prasad  (9 years ago)

3. Always divisable by 3

Is this solution Helpfull? Yes (2) | No (5)  

vinay  (9 years ago)

All the above.... Test for 1000,1001 or any number pair

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MOHIT VERMA  (9 years ago)

answer is d.(all of the above)
7^3>6^3>5^3
343-256=127
127-1=126 which is divisible by all.

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KANCHAN MALA  (9 years ago)

Akash kumar D-1 = 3*x*Y - 1
and D= 3XY

Is this solution Helpfull? Yes (1) | No (1)  

Mithun Kumar Kesari  (9 years ago)

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.

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