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Source: — Data Sufficiency |
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by eagleeye » Sat Jul 21, 2012 7:32 pm
confuse mind wrote:Is y an integer?

1. y^3 is an integer
2. 3y is an integer
[spoiler]


IMO - C[/spoiler]

1. Let y^3 = m ( where m is an integer)
y = m^1/3. If m is 1, y is integer. If m= 2, it isn't. Insufficient.

2. Let 3y=n (n is an integer)
=> y=n/3.
If n=1, y=1/3
If n=3, y=1. Insufficient.

Together, equating the 2 values of y.
n/3 = m^(1/3)
=> m = (n/3)^3.
Since m is an integer, n must be a multiple of 3.
Hence n=3k (where k is another integer).

So y=n/3=3k/3 =k. Hence y is an integer. Sufficient.

C is correct. :)
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by theCEO » Sat Jul 21, 2012 9:59 pm
eagleeye wrote:
confuse mind wrote:Is y an integer?

1. y^3 is an integer
2. 3y is an integer
[spoiler]


IMO - C[/spoiler]

1. Let y^3 = m ( where m is an integer)
y = m^1/3. If m is 1, y is integer. If m= 2, it isn't. Insufficient.

2. Let 3y=n (n is an integer)
=> y=n/3.
If n=1, y=1/3
If n=3, y=1. Insufficient.

Together, equating the 2 values of y.
n/3 = m^(1/3)
=> m = (n/3)^3.
Since m is an integer, n must be a multiple of 3.
Hence n=3k (where k is another integer).

So y=n/3=3k/3 =k. Hence y is an integer. Sufficient.

C is correct. :)
eagleeye,

Could you explain why 1) is not sufficient?

The question ask is y an integer if y^3 is an integer?

y^3 is integer..
therefore y x y x y = integer
noninteger * noninteger * noninteger = non-integer
integer * integer * integer = integer

Isn't this satisfied?

According to this question, m could never be 2 :)
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by eagleeye » Sat Jul 21, 2012 10:08 pm
theCEO wrote:
eagleeye,

Could you explain why 1) is not sufficient?

The question ask is y an integer if y^3 is an integer?

y^3 is integer..
therefore y x y x y = integer
noninteger * noninteger * noninteger = non-integer
integer * integer * integer = integer

Isn't this satisfied?
Integer* Integer * Integer = Integer (always).

But Non-integer * Non-integer * Non-integer may be an integer, may be not.

Example :

cube root (2) * cube root (2) * cube root (2) = cube root (2^3) = cube root (8) = 2
Hence m can be 2 when y is 2^(1/3). That's why 1 is insufficient.
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by theCEO » Sat Jul 21, 2012 10:10 pm
eagleeye wrote:
theCEO wrote:
eagleeye,

Could you explain why 1) is not sufficient?

The question ask is y an integer if y^3 is an integer?

y^3 is integer..
therefore y x y x y = integer
noninteger * noninteger * noninteger = non-integer
integer * integer * integer = integer

Isn't this satisfied?
Integer* Integer * Integer = Integer (always).

But Non-integer * Non-integer * Non-integer may be an integer, may be not.

Example :

cube root (2) * cube root (2) * cube root (2) = cube root (2^3) = cube root (8) = 2
Hence m can be 2 when y is 2^(1/3). That's why 1 is insufficient.
Perfect! Thanks for the explanation
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