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If x and y are positive integers, is x^17 - 2*y^3 odd? 15 Dec 2020, 01:15
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If x and y are positive integers, is \(x^{17} - 2*y^3\) odd?

(1) The median of x consecutive even integers is an even integer.
(2) 2x is divisible by twice as many positive integers as x


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If x and y are positive integers, is x^17 - 2*y^3 odd? 15 Dec 2020, 02:47
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The question stem has two positive integers x and y. x^17-2y^3 has already one term 2y^3 as even so we need to find whther x is odd.

Statement 1: median of x consecutive even numbers is even.

This condition is satisfied only if x is odd.

Let's take 2, 4 and 6. X is odd here and median is 4 which is even.

Let's take 2,4,6 and 8 now median is 5 which is odd.

So we can infer x is odd for median to be even. SUFFICIENT

Statement 2: this says 2x is divisible by twice the number of x positive integers.

This can happen only if x = 1 where 2*1 has two factors.

X is odd. SUFFICIENT.

Answer D

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If x and y are positive integers, is x^17 - 2*y^3 odd? 15 Dec 2020, 03:52
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Bunuel
If x and y are positive integers, is \(x^{17} - 2*y^3\) odd?

(1) The median of x consecutive even integers is an even integer.
(2) 2x is divisible by twice as many positive integers as x


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Basically we need to determine if 'X' is odd. We need to get an answer in the form of Yes or No.

(1) The median of x consecutive even integers is an even integer.
Example : 0,2,4,6 (median is 3 and x is even)
0,2,4 (median is even and x is odd)
(A) is sufficient - we know that x is ODD for sure

(2) 2x is divisible by twice as many positive integers as x
Consider example of x=3 or x=5
Here, the odd numbers satisfy the above criteria.
(B) is sufficient.

IMO D is the correct answer.
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If x and y are positive integers, is x^17 - 2*y^3 odd? 15 Dec 2020, 05:27
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Quote:
If x and y are positive integers, is \(x^{17} - 2*y^3\) odd?

(1) The median of x consecutive even integers is an even integer.
(2) 2x is divisible by twice as many positive integers as x
Show more

Odd-Even = Odd

From the stem, we know 2*\(y^3\) = Even. Therefore \(x^17\) needs to be odd.
Hence, the only thing we need to know is Whether x is odd.

1) The median of x consecutive even integers is an even integer.

\(\frac{(Odd times Even) }{ Odd }\) = Odd eg. \(\frac{(2+4+6)}{3}\) = 4
\(\frac{(Even times Even) }{ Even }\) = \(\frac{(2+4+6+8)}{4}\) = 5

So, in order for x consecutive even integers to be even, x has to be odd

Sufficient

2) 2x is divisible by twice as many positive integers as x

It is basically saying that the number of factors of 2x is twice the value of x
eg. x=1. Therefore, 2x = 2
No. of factors of \(2^1\) = 1+1 = 2 ( 1 and 2)
2x is divisible by twice as many integers as x i.e 2x is divisible by 2 integers. 1 and 2 both

Take x = 3. Therefore 2x = 6
No. of factors of \(2^1\)*\(3^1\) = (1+1)*(1+1) = 2*2 = 4 (1, 2, 3, 6)
2x should be divisible by twice x i.e 6. But, this is not possible.
Hence, x = 1. Odd
sufficient

IMO (D)
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If x and y are positive integers, is x^17 - 2*y^3 odd? 15 Dec 2020, 10:56
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If x and y are positive integers, is \(x^{17} - 2*y^3\) odd?

(1) The median of x consecutive even integers is an even integer.
(2) 2x is divisible by twice as many positive integers as x


to find out x^{17} - 2*y^3 is odd
in other words we have to find out is a - 2(b) Odd?
or
Is a - even = odd?
or
Is a = odd? => x ^ 17 = odd?
or
Is x = odd?

1)consecutive even integers -->
If no of integers are odd e.g. 2, 4, 6 => median = 4, an even no.
If no of integers are even e.g. 2, 4, 6, 8 => Median is 4+6/ 2 = 5, an odd no.

As from above x is odd. => C, E gone

2) 2x is divisible by twice as many positive integers as x

Point to remember: total number of Factors for any number = (a+1) (b+1) (c+1) when the number is in the form of (x^a)(y^b)(z^c) | X, Y, Z are prime factors and a, b, c are their corresponding powers

even

lets x = 4 => 2x = 8
Factors of x or (2)^2 = 2+1 = 3
Factors of 2x or (2)^3 = 3+1 = 4
4 is not twice of 3

Odd

Lets x = 5=> 2x =10
Factors of x or 5 = 1+1 = 2
Factors of 2x or 10 or (5^1)(2^1) = (1+1)(1+1) =4

Seems right
Let me check one more example

Lets x = 9 => 2x = 18
Factors of x or 9 or 3^2 = 2+1= 3
Factors of 2x or 18 or (3^2)(2^1)= (2+1)(1+1)=3*2= 6
This also works..

Basically whenever there is only one multiple of 2, there would be (2^1) = Multiple of (1+1) factors or in other terms twice the factor
If there is more than one multiple of 2 [one introduced by 2 and atleast one introduced by the original number by virtue of being an Even integer.. for instance minimum (2^2) there would be (2+1) or three times extra factor (in this case)


Both are sufficient
IMO [D]
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If x and y are positive integers, is x^17 - 2*y^3 odd? 15 Dec 2020, 14:42
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Even - Odd = Odd
Odd - Even = Odd

Since y is multiplied by 2, we know that y is even, so we need to find whether x is odd.

(1) The median of x consecutive even integers is an even integer.

when x is even , x=4 , consecutive even integers for example = 2,4,6,8
Then the medium is 5 which is Odd

when x is odd, x=5, consecutive even integers are 2,4,6,8 and 10, then the median is even.
so we know that x is odd

Option A - Sufficient.

(2) 2x is divisible by twice as many positive integers as x

Consider x is even, x=4, then we have 2x = 8
x = 4 = \(2^2\) = number of factors = 3
2x = 8 = \(2^3\) = number of factors = 4

The 2x is not divisible by twice as many positive integers as x

Lets consider x as odd, x=5, then 2x=10
x=5, \(5^1\) = number of factors = 2
x=10 , \(2^1\) * \(5^1\) = number of factors = 4

The 2x is divisible by twice as many positive integers as x when x is odd

Option B - Sufficient

Ans: D
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If x and y are positive integers, is x^17 - 2*y^3 odd? 15 Sep 2025, 18:06
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