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Bunuel
If x is a positive integer, is x^2 + 1 an odd number?

(1) x is the smallest integer that is divisible by all integers from 11 to 15, inclusive.
(2) 3^x is an odd number.


So \(x\geq{1}\)
Thus the answer will depend on x. If x is odd, x^2+1 is even, otherwise odd.


(1) x is the smallest integer that is divisible by all integers from 11 to 15, inclusive.
x is divisible by 11, 12, 13, 14, 15.
Since x is divisible by EVEN numbers, x is even.
Suff

(2) 3^x is an odd number.
x could be odd or even, as 3^x will always be odd.
Insuff

A
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Bunuel
If x is a positive integer, is x^2 + 1 an odd number?

(1) x is the smallest integer that is divisible by all integers from 11 to 15, inclusive.
(2) 3^x is an odd number.


x>=1
x2+1=odd?
x2+odd=odd, when x=even

(1) sufic
x/{11,12...15}=integer
x=even

(2) insufic
3^x=odd, then x = {odd or even}

(A)
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Bunuel
If x is a positive integer, is x^2 + 1 an odd number?

(1) x is the smallest integer that is divisible by all integers from 11 to 15, inclusive.
(2) 3^x is an odd number.


Project DS Butler Data Sufficiency (DS3)


For DS butler Questions Click Here

Is \(x^2 + 1\) odd?
Is \(x^2\) even?
Is \(x\) even?

Statement 1:
"The smallest integer" reveals there is only one such value for x, since there is only one value for x this is automatically sufficient.

Statement 2:
No matter how high you raise the power of 3, \(3^x\) is always going to odd, so this doesn't tell us anything about x. Insufficient.

Ans: A
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If x is a positive integer, is x^2 + 1 an odd number?

(1) x is the smallest integer that is divisible by all integers from 11 to 15, inclusive.
(2) 3^x is an odd number.


Statement 1:

x = 3*4*5**7*11*13

x^2 + 1, identifiable

Sufficient

Statement 2:

for any positive value of x , 3^x = odd

Notsufficient

Answer A
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If x is a positive integer, is x^2 + 1 an odd number?

x^2 + 1 will be an odd number only when x is even.


(1) x is the smallest integer that is divisible by all integers from 11 to 15, inclusive.
Therefore, x is even.
Sufficient.

(2) 3^x is an odd number.
Therefore, x can be 1, 2, 3 etc.

So x can be even or odd.

Not sufficient.

Option A
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