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Magoosh Brain Twister: Nasty Squares and Cubes

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Good luck with this challenging numeric entry problem!

The square root of integer x is equal to the sum of the cubes of y and z. If the squares of y and z are each less than 10, what is the greatest possible value of x?

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Don’t forget to check back on Thursday for the answer and explanation. 🙂

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38 Responses to Magoosh Brain Twister: Nasty Squares and Cubes

  1. Oxana October 9, 2014 at 3:01 am #

    I’ve got it as 3982

    • Chris Lele
      Chris Lele October 9, 2014 at 12:05 pm #

      Very close! The explanation is out today, so take another shot but if you can’t figure it out, hopefully that will help 🙂

  2. Abunaser October 8, 2014 at 11:53 pm #

    In the question, it did not mention that y and an z must be an integer. So, I am assuming value of y and z square could be any number less than 10 in which case, each of the square of y and and z could possibly be 9.9999999 ( as you can see the digit of 9 after decimal could go towards infinity because there are infinite numbers between two integers) at most. So, the value of y and z square approximately is 10. The value of y and z is very close to the root square of 10. By plugging in the value of y and z square and value of y and z to solve for x, I get x=4000(upper limit for the value of integer x).any integer less than 4000 is 3999. My answer turns out to be 3999.

    My solution

    X=(y^3+z^3)^2
    X=y^6+2y^ (2) z^(3)+ z^(6)
    X=4000 _ ( x must be little <4000 as an integer)
    So, x is 3999 at most.

    Answer 3999.

    • Chris Lele
      Chris Lele October 9, 2014 at 12:06 pm #

      You got it!

      • Abunaser October 9, 2014 at 1:12 pm #

        Hi Chris. Thank you very much for posting some of the most challenging math problems for GRE. I have one relevant question related to this question. If the question was worded little differently and asked to round up answer to the nearest whole number,would the answer be 4000 because the value of x come out to be very close to 4000 ( possibly 3999.99 because y and z square could be 9.999999999999 or very close to 10 for the purpose of approximation. Please reply.

        • Chris Lele
          Chris Lele October 10, 2014 at 10:23 am #

          Exactly! If the question had asked us to round up to the closest integer, the answer would have been 4,000 (as you point out 9.999999 will do the trick :)). Good question!

        • Anh Le June 19, 2015 at 10:56 am #

          I think 3999 is good enough because x is an interger

  3. SAAKSHY October 8, 2014 at 9:06 pm #

    The answer should be 2916.

    • Chris Lele
      Chris Lele October 9, 2014 at 12:06 pm #

      Remember: only ‘x’ has to be an integer.

      Hope that helps 🙂

  4. Sanjay October 8, 2014 at 10:45 am #

    sqrt(x) = y^3 + z^3. Given: y^2 < 10 and z^2 < 10.

    sqrt (10) = 3.16 (approx). Considering y and z are not integers as nothing is told about x and y. y^3 < 31.55 approx and similarly z^3 < 31.55. Hence sqrt(x) < 31.55 + 31.55

    So, x < (63.1)^2 = 3981.61 . As x is given to be an integer and we are asked the greatest possible value of x as an integer which is 3981 probably.

    Hence the answer is 3981.

    • Chris Lele
      Chris Lele October 9, 2014 at 12:54 pm #

      Very close. But y and z can be even greater than you assume–albeit by very little. The explanation is up today, so you can have another go at it and then check your work 🙂

  5. Harish October 7, 2014 at 9:05 am #

    If y, z need not be integers then y<sqrt(10), z<sqrt(10). y^3 + z^3 < 2*10^(3/2) ,
    sqrt(x)<2*10^(3/2). We are going for greatest value, hence negative values can be ignored so x < 4000. Greatest integer x= 3999.

    Is 3999 correct.

    • Chris Lele
      Chris Lele October 9, 2014 at 12:54 pm #

      Got it!

  6. Amit Poddar October 7, 2014 at 5:08 am #

    y^2 y -(10^1/2)
    z^2 z -(10^1/2)

    y^3 -(10^3/2)
    z^3 -(10^3/2)

    y^3 + z^3 is maximized when both are positive

    y^3 + z^3 < 2*(10^3/2)
    x^1/2 < 2*(10^3/2)
    x < 2*(10^3)
    x < 2000

    Answer: 2000

    • Chris Lele
      Chris Lele October 9, 2014 at 12:55 pm #

      I think you missed something small and forgot to double 2,000, which would then be very close to the answer. The explanation post should be up now 🙂

  7. Astrid October 7, 2014 at 3:20 am #

    Is it 0?

    • Chris Lele
      Chris Lele October 9, 2014 at 12:56 pm #

      Not quite–keep trying :). Also, the explanation post is up today!

  8. ramya October 6, 2014 at 11:45 pm #

    X= 2916…

    • Chris Lele
      Chris Lele October 9, 2014 at 12:56 pm #

      You’re off by a little, but you seem to be on the right path :). Remember, the problem only states that ‘x’ has to be an integer.

  9. Bilal Saeed October 6, 2014 at 11:17 pm #

    I guess 3999.

    • Chris Lele
      Chris Lele October 9, 2014 at 12:57 pm #

      You got it!

  10. Revathy October 6, 2014 at 10:50 pm #

    I think answer could be X= 2916..

    • Chris Lele
      Chris Lele October 9, 2014 at 1:02 pm #

      The problem only states that ‘x’ has to be an integer 🙂

  11. sanjay October 6, 2014 at 8:45 pm #

    I got 2916 as answer

    • Chris Lele
      Chris Lele October 9, 2014 at 1:03 pm #

      I think you did what a lot of people did, and that was assume that ‘y’ and ‘z’ are integers. Try it again without that assumption and see what you get it 🙂

  12. Dmytro October 6, 2014 at 3:24 pm #

    So long as both Z and Y are integers, the greatest possible integer that yields the number less than 10 when squared is 3. So, if Z and Y both equal 3, their cubes are 27 that gives the sum of 54. 54 raised to the power of two is 2916.

    • Oxana October 9, 2014 at 2:59 am #

      only x is defined as an integer

    • Chris Lele
      Chris Lele October 9, 2014 at 1:03 pm #

      Oxana is right. Take another shot at it (the explanation should be up :)).

  13. bandhanbio October 6, 2014 at 12:31 pm #

    2916

    • Chris Lele
      Chris Lele October 9, 2014 at 1:04 pm #

      Only ‘x’ has to be an integer. See if you can rework the problem 🙂

  14. vignesh October 6, 2014 at 11:46 am #

    1458

    • Chris Lele
      Chris Lele October 9, 2014 at 1:05 pm #

      Hmm…I think you were the only one who got that. How did you arrive at this number?

  15. GRE_HUNTER October 6, 2014 at 10:49 am #

    Sarvana , yes u are wrong. The correct answer is 3999.

    • Chris Lele
      Chris Lele October 9, 2014 at 1:05 pm #

      That is correct 🙂

  16. anisha October 6, 2014 at 10:38 am #

    Are y and z integer?

    • Chris Lele
      Chris Lele October 9, 2014 at 1:06 pm #

      If the problem doesn’t state that they are, then you cannot assume so. Take a stab at the problem, knowing that ‘y’ and ‘z’ do not have to be integers 🙂

  17. Saravana October 6, 2014 at 9:38 am #

    I think the answer may be 1225!! Tell me I’m wrong…! 😛

    • Chris Lele
      Chris Lele October 9, 2014 at 1:07 pm #

      Hm…you are off a little :). How did you get 1225?


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