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Magoosh Brain Twister: A Known Unknown



Twist your brain with this challenging GRE problem! Then, leave a comment with your work and best guess. Good luck!

x^n + x^n + x^n = x^(n + 1)

Which of the following must be true?

I. n = 2
II. x = 3
III. n^x < x^n

(A) I only

(B) II and III

(C) I and III

(D) II only

(E) None of the above

See you on Friday for the answer and explanation. 🙂

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20 Responses to Magoosh Brain Twister: A Known Unknown

  1. Vishal September 17, 2016 at 12:08 am #

    X=3 and n has infinite solution!

  2. RAFI BERCHA June 9, 2015 at 4:40 am #

    D is the right option since it satisfies the given equation and all the other option don’t so rest of the options are wrong.

    • Chris Lele
      Chris Lele June 9, 2015 at 3:28 pm #

      Yes, (D) is the answer 🙂

  3. RAFI BERCHA June 9, 2015 at 4:37 am #

    Answer is D .. the option satisties the given equation and all the other options does not.

  4. Ishan May 31, 2015 at 10:24 pm #

    x^n+x^n+x^n = x^(n+1)

    Solving this,

    3 * x^n = x^n * x
    => x = 3.

    Notice that we can cancel x^n term, which means for any value of n, the equation will hold. So answer has to be (D). We see something similar when two linear equations have infinite solutions.

  5. H-MAN May 30, 2015 at 8:45 am #

    Answer should be D. For X=3, for all values of “n” equation balances out

  6. Manjinder Singh May 29, 2015 at 12:14 am #

    x can be both zero or 3.
    so none of the above is must.

    answer would be E.

  7. Manjinder Singh May 29, 2015 at 12:01 am #

    E) none of the above

    • Anh Le June 9, 2015 at 11:00 pm #

      x cannot be 0

      If x=0 and n=0, the left side will equal 3, but the right side equals 0. This fail to satisfy the equation

      • Manjinder Singh October 15, 2015 at 2:19 am #

        @Ahh Le

        x and n can not be equal to 0 simultaneously because that would form a indeterminate form
        that is “zero raised to power zero”.
        which is not defined in world of mathematics

        this question had some typos so it was confusing 🙂

  8. Vinod May 28, 2015 at 8:36 am #

    Rearranging, we get


    For this to be true, one of the following conditions must be true:

    –> x=0, n can be any real number
    –> x=3

    However, the questions states that which of the following must be true, so while any of the options given can be true for the given statement to be valid, there is a possibility that the other condition is what is actually making the statement valid.

    So there is no surety about which condition must be true.

    Hence, the answer is (E).

  9. WALEED ELGRAWANI May 27, 2015 at 11:38 am #

    the right answer is B both 2 and 3 is right ….
    using the trying number technique by trying both the even and odd possibility
    the first possibility 1* N cannot equal 2 as when u try 4 as a base for the square 2 both sides of the equation will not be equal .
    on the other hand using the same technique of trying number u will find always that 2 and 3 must be true

    • Ashley Long May 28, 2015 at 7:32 am #

      III. Cannot be right.

      The answer is (D). Not (B).

      Here is why…

      If x=3 and n can equal any number,


      Now for part III.

      n^x < x^n
      3^3 < 3^3
      9 < 9 (This is not right, 9 cannot be less than 9)

      Therefore…the correct answer is (D.)

  10. Harry May 27, 2015 at 5:50 am #

    The answer is II (x = 3).

    x^n + x^n + x^n = x^(n + 1)
    => 3 * x^n = x * x^n
    => x = 3 (x^n at both sides of the equation is cancelled out)

  11. Nalin May 27, 2015 at 5:47 am #

    Answer is (D) : II only

    x^n + x^n + x^n = x^(n+1)
    => 3.x^n = (x^n).(x)
    =>3 = x

    also, 3^n and n^3 will be equal for n=3; therefore III is wrong

  12. sriram May 26, 2015 at 8:45 pm #

    the eqn can be written as 3x^n =x^(n+1)

    x=3 satisfies he eqn for any power for ex: x=10 gives 3(3^10)=3^11–>3^11=3^11 same for negative and fractional powers

    n=2 need not be true for all values of x, but n=2 goes well for x=3..hence n=2 is not correct

    3rd condition works for x=3. but not for different values of x and n

    hence d is the answer

  13. Nimish May 26, 2015 at 7:57 am #

    x^n + x^n + x^n = x^(n + 1)
    3*(x^n) = x^(n+1)

    n=real number

    for n=3, III is incorrect

    D option

  14. Premanand Ramesh May 26, 2015 at 6:28 am #

    I would go with D. n =2 doesn’t guarantee a 0 on the RHS of the equation. Similarly, one cannot draw a determinate conclusion about n and x in the third option. But, x=3 guarantees an RHS=0. Hence the guess

  15. sahil May 26, 2015 at 1:41 am #

    Answer is “D”

  16. Shashank May 25, 2015 at 9:29 pm #

    (E) none of the above.
    Since I and II are true.

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