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If y is an integer, is y^3 divisible by 8? (1) y is even. (2) y^3 - y

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Re: If y is an integer, is y^3 divisible by 8? (1) y is even. (2) y^3 - y [#permalink]
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If y is an integer, is y^3 divisible by 8?

(1) y is even.
(2) y^3 - y is even

SOLUTION - (A)

EXPLAINED

statement 1 - y is when
Let y = 2x
And hence (2x)^3 = 8x^3 , which is divisible by 8
SUFFICIENT

Now we are left with options A and D

statement 2 (y^3-y) is even
When y is odd, (Odd no.)^3 - odd = even
When y is even, (even no.)^3 - even = even
Hence, it cannot be determined if 'y' is even or odd.
NOT SUFFICIENT

THEREFORE, option D is eliminated

Answer - A

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Re: If y is an integer, is y^3 divisible by 8? (1) y is even. (2) y^3 - y [#permalink]
Bunuel wrote:
If y is an integer, is y^3 divisible by 8?

(1) y is even.
(2) y^3 - y is even.

Project PS Butler

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target check y^3/8 is divisible
possible only when y is even
#1
y is even
so sufficient
#2
y^3-y is even
y*(y^2-1) is even
y can be odd or even ; hence insufficient
OPTION A
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Re: If y is an integer, is y^3 divisible by 8? (1) y is even. (2) y^3 - y [#permalink]
1
Kudos
y is an integer.
If $$y^3$$ has to be divisible by 8, then y has to be even.
The question can be rephrased as “Is y even?”. This is because the cubes of all even integers will definitely be divisible by 8.

From statement I alone, y is even. This is sufficient to answer the question.
Answer options B, C and E can be eliminated. Possible answer options are A or D.

From statement II alone, $$y^3$$ – y is even. Let us break the LHS down into a simpler expression.

$$y^3$$ – y = even. Taking y common, LHS becomes,

y($$y^2$$-1) = even. We can consider 2 cases now.

If y = even, $$y^2$$ = even and $$y^2$$-1 = odd; y($$y^2$$-1) = even * odd = even. Is y even? YES.
If y = odd, $$y^2$$ = odd and $$y^2$$-1 = even; y($$y^2$$-1) = odd * even = even. Is y even? NO.

Statement II alone is insufficient to find a definite YES or NO. Answer option D can be eliminated.

The correct answer option is A.

Breaking down the question stem always impels you forward by a few steps in your solution. So, practice this technique as much as you can when you solve DS questions.
Hope that helps!
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Re: If y is an integer, is y^3 divisible by 8? (1) y is even. (2) y^3 - y [#permalink]
A straightforward question.

y is an integer and we need to find if y^3 is 8k (k is any +ve integer)

Statement A -> y is even. so y = 2m (m is any +ve integer)
The statement is sufficient

Statement B -> y^3-y = 2m (m is any +ve integer)

y(y^2-1) = 2m
so y is even.
But same cannot be dictated for y^2-1.
Hence the statement is insufficient.

Answer - A
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Re: If y is an integer, is y^3 divisible by 8? (1) y is even. (2) y^3 - y [#permalink]
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Re: If y is an integer, is y^3 divisible by 8? (1) y is even. (2) y^3 - y [#permalink]
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