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Do not try to do it using the decomposition in factors: that would be a non-sense, as you are told that 152=X^3+Y^3 (and not X^3*Y^3). Just do the cube of the first integers: 1^3=1 ------> 2^3=8 ------> 3^3=27 ------> 4^3=84 ------> 5^3=125 ------> when reaching this point, we can see that 125+27=152. Therefore, X=3 and Y=5. And the product of X*Y=15. Solution B.

More:"All I wish someone had told me about GMAT beforehand" There are many things you want to know before doing the GMAT exam (how is exam day, what to expect, how to think, to do's...), and you have them in this blog, in a simple way

Re: The number 152 is equal to the sum of the cubes of two integ [#permalink]
27 Feb 2013, 01:06

1

This post received KUDOS

Expert's post

johnwesley wrote:

The number 152 is equal to the sum of the cubes of two integers. What is the product of those integers?

A) 8 B) 15 C) 21 D) 27 E) 39

Do not try to do it using the decomposition in factors: that would be a non-sense, as you are told that 152=X^3+Y^3 (and not X^3*Y^3). Just do the cube of the first integers: 1^3=1 ------> 2^3=8 ------> 3^3=27 ------> 4^3=84 ------> 5^3=125 ------> when reaching this point, we can see that 125+27=152. Therefore, X=3 and Y=5. And the product of X*Y=15. Solution B.]

Apart from 3 and 4, -4 and 6 also satisfy the given condition: (-4)^3 + 6^3 =152, so the product could also be -4*6 = -24. _________________

More:"All I wish someone had told me about GMAT beforehand" There are many things you want to know before doing the GMAT exam (how is exam day, what to expect, how to think, to do's...), and you have them in this blog, in a simple way

Re: The number 152 is equal to the sum of the cubes of two integ [#permalink]
27 Feb 2013, 20:26

1

This post received KUDOS

Expert's post

johnwesley wrote:

The number 152 is equal to the sum of the cubes of two integers. What is the product of those integers?

A) 8 B) 15 C) 21 D) 27 E) 39

Do not try to do it using the decomposition in factors: that would be a non-sense, as you are told that 152=X^3+Y^3 (and not X^3*Y^3). Just do the cube of the first integers: 1^3=1 ------> 2^3=8 ------> 3^3=27 ------> 4^3=84 ------> 5^3=125 ------> when reaching this point, we can see that 125+27=152. Therefore, X=3 and Y=5. And the product of X*Y=15. Solution B.

Actually, decomposition into factors can easily give you the answer here. You should just do the decomposition of the right thing i.e. the options since they represent the product of those integers.

Since the sum of cubes is 152, the numbers cannot be larger than 5 since 6^3 itself is 216.

Re: The number 152 is equal to the sum of the cubes of two integ [#permalink]
28 Feb 2013, 13:21

VeritasPrepKarishma wrote:

johnwesley wrote:

The number 152 is equal to the sum of the cubes of two integers. What is the product of those integers?

A) 8 B) 15 C) 21 D) 27 E) 39

Do not try to do it using the decomposition in factors: that would be a non-sense, as you are told that 152=X^3+Y^3 (and not X^3*Y^3). Just do the cube of the first integers: 1^3=1 ------> 2^3=8 ------> 3^3=27 ------> 4^3=84 ------> 5^3=125 ------> when reaching this point, we can see that 125+27=152. Therefore, X=3 and Y=5. And the product of X*Y=15. Solution B.

Actually, decomposition into factors can easily give you the answer here. You should just do the decomposition of the right thing i.e. the options since they represent the product of those integers.

Since the sum of cubes is 152, the numbers cannot be larger than 5 since 6^3 itself is 216.

21, 27, 39 - The factors are too large so ignore

8 - (2, 4) Does not satisfy

15 - (3, 5) Yes. 3^3 + 5^3 = 152 - Answer

You nailed it Karishma, that was exactly my approach. Since 6^3 surpasses 152, I sticked with an answer choice that had a factor of <6. Then with a little math I discarded 8, having 15 alone as my answer choice. _________________

MV "Better to fight for something than live for nothing.” ― George S. Patton Jr

Re: The number 152 is equal to the sum of the cubes of two integ [#permalink]
19 Jul 2014, 04:13

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