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I’ve got it as 3982

Very close! The explanation is out today, so take another shot but if you can’t figure it out, hopefully that will help ๐

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.

You got it!

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.

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!

I think 3999 is good enough because x is an interger

The answer should be 2916.

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

Hope that helps ๐

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.

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 ๐

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.

Got it!

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

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 ๐

Is it 0?

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

X= 2916…

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.

I guess 3999.

You got it!

I think answer could be X= 2916..

The problem only states that ‘x’ has to be an integer ๐

I got 2916 as answer

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 ๐

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.

only x is defined as an integer

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

2916

Only ‘x’ has to be an integer. See if you can rework the problem ๐

1458

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

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

That is correct ๐

Are y and z integer?

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 ๐

I think the answer may be 1225!! Tell me I’m wrong…! ๐

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