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Given:
If m and n are prime numbers, is (m - 11)(2 - n) an odd number ?
(1) m > 11
(2) n < 3

Evaluation:
1) m> 11:
We can plug in a few values: 13, 17, 19. The expression (m-11) becomes even.
Now, we don't know anything about n.
It can be 2 (the only even prime number), or 3, or 11, or anything. Let's check that.
2-2 = 0 So in that case, the whole expression becomes 0 (even). The expression becomes even.
2-3 = 1 So in that case, the whole expression remains even.
2-5 = 3 So in that case, the whole expression remains even. (as even x odd) = even.
So in all scenarios, we can see that the expression gives us an even number (Sufficient)

2) n < 3:
In this case, since it's already given that n is prime, we have only one value for n, that is 2.
The whole expression becomes 0, which is even.
In this case also, we don't know the value of m, but in all cases, the value remains even.
(Sufficient)

So, I will go with option D.
Each statement alone is sufficient.
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If m and n are prime numbers, is (m - 11)(2 - n) an odd number?

(1) m > 11
(2) n < 3
We know that 2 is the only even prime number.
X*Y = odd, if both X and Y are odd.
X*Y = even, if any of X and Y or Both are even.

(m-11) = odd, if m is even, i.e. 2
(m-11)= even, if m is any prime number other than 2.
(2-n)= 0 that is even, if n=2
(2-n) = odd, if n is any prime number other than 2



(1) m>11 means m is odd number, hence (m-11) is even number. The expression is an even number and the answer is NO. --> SUFFICIENT

(2) n<3 means n=2. So (n-2) = 0. The expression equals to zero, which is even number and the answer is NO. --> SUFFICIENT

IMO: D
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If m and n are prime numbers, is (m - 11)(2 - n) an odd number ?

(1) m > 11
m can be 13, 17, 23, . . .
hence (m-11) will be even number
and even*odd/Even is always even
Hence the Product of (m - 11)(2 - n) will not be odd
Statement 1 Sufficient

(2) n < 3
n is a prime number and it is also less than 3, the only possible value of n is 2
(n-2)=0
hence the product of (m - 11)(2 - n) will be zero and it will not be odd
Statement 2 Sufficient

Option D
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(1) m > 11:
Since m is a prime number and is greater than 11 which means m is a odd number as apart from 2 there cannot be a even positive integer.
Now \((m-11)\) is odd - odd which is even irrespective of (2-n)

Statement (1) alone is sufficient to answer the question with a "No."

(2) n < 3:
Since n is a prime number and is less than 3, the only possible value for n is 2 (since prime numbers are greater than 1 and the only primes less than 3 are 2). Now, we need to consider (m-11) and (2-n):

If n = 2, then \((2-n) = (2-2) = 0\).

Regardless of the value of m, when (2 - n) is 0, the product (m - 11)(2 - n) will always be 0, which is an even number.

Statement (2) alone is sufficient to answer the question with a "No."

Since both statements are individually sufficient to determine the answer, the answer is option (D)
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Bunuel
If m and n are prime numbers, is (m - 11)(2 - n) an odd number ?

(1) m > 11
(2) n < 3

m & n could be equal to 2,3,5,7, or any other prime numbers.

From Statement 1,we get

m = 13/17/19/.. and so on. Therefore, m is an odd prime number.
Hence, m-11 = odd - odd = even

This implies that (m - 11)*(2 - n) = (Odd- 11)*(2- n) =Even*(2- n)= Even

Statement 1 is sufficient.

From Statement 2, we get

n<3
Therefore, only possible value of n is 2.
(m - 11)*(2 - n) = (m - 11)*(2 - 2) = (m- 11)* 0 = 0 = Even

Statement 2 is sufficient.

IMO OA should be D.
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