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

Question

If y is an integer, is y^3 divisible by 8?

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

GMATinsight · Answer

Question: If y is an integer, is y^3 divisible by 8?

Statement 1 ) y is even.

y = 0 (even), y^3=0 is divisible by 8
y = 2 (even), y^3=8 is divisible by 8
y = -2 (even), y^3=-8 is divisible by 8

Answer is always YES hence

SUFFICIENT

Statement 2 ) y^3 - y is even.

i.e. y(y^-1) = (y-1)*y*(y+1) = even

But since product of three consecutive integers (y-1)*y*(y+1) is ALWAYS EVEN therefore y may be even as well odd

@y= 1 y^3=1 which is NOT divisible by 8 but
@y = 2 (even), y^3=8 is divisible by 8

NOT SUFFICIENT

ANswer: Option A

GMATWhizTeam · Answer

Solution 
 Step 1: Analyse Question Stem 
 • y is an integer.
 o  y^3  also be an integer.  
We need to find if  y^3  is divisible by 8.
 • If  y^3  is divisible by 8, this means  y^3  must be in the form of  y^3 = 8*k^3 , where k is an integer.
• Since,  y^3 = 2^3*k^3 ⟹ y = 2*k 
• So,  y^3  will be divisible by 8 if y is an even number. 
Basically, we need to find the even odd nature of y.

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE 
 Statement 1:  y is even.
 • Since y is even so we can conclude that  y^3  is divisible by 8. 
Hence, statement 1 is sufficient and we can eliminate answer Options B, C and E.
 Statement 2:   y^3 - y  is even.
 • We know that if y is even then  y^3  will be even and if y is odd then  y^3  will be odd.
• So, according to this statement, there can be two cases:
 o Case 1: If  y  and  y^3  are even, then  even – even = even 
  In this case  y^3  will be divisible by 8 
o Case 2: If y and  y^3  are odd, then  odd – odd = even 
  In this  y^3  will not be divisible by 8.  
• Here we are getting contradictory results. 
Hence, statement 2 is NOT sufficient.
Thus, the correct answer is  Option A.

sukriti16 · Answer

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

Posted from my mobile device

Archit3110 · Answer

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

CrackverbalGMAT · Answer

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!

sambitspm · Answer

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

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