Around the World in 80 Questions (Day 3): If m and n are prime numbers

It is asked that if product of two expressions is odd number or not.
A product can be odd only when all of the numbers involved in the product are odd.
Consider individual expressions
1) m-11 can only be odd number if m is even. The only possibility of m to be even is m=2 and prime number
2) 2-n can only be odd number if n is odd. Here there since n is a prime number there are various possibilities. We just need to make sure n!=2

Statement 1: m>11, m is definitely an odd number since no prime number >11 is even. Hence the expression (m-11) is even since (odd-odd) is even.
We know that any number*2 is even. The product is even number. Hence statement 1 is sufficient.

Statement 2: n>3, m=2. Hence the expression 2-n=0. The product is 0 not an odd number. Hence statement 2 is sufficient.

Answer D

Statement 1- If m> 11, m must be odd. So, odd - odd = even, so, (m-11) * (2-n) will always be even regardless of the value of n. So, sufficient.

Statement 2- If n<3, it can only be 2, so the product of (m-11) * (2-n) will always be 0, so even. So, sufficient.

Since each statement is individually sufficient, Option D is the answer.

D is the answer.

1 alone is sufficient since m can take only prime numbers, m- 11 will always be even and so the product will also be even.
2 alone is sufficient since n can take the value of 2 so the product of 0*(m-11) is always 0 which is even.

1) m>11 => m must be odd => m-11 must be even => (m-11)(2-n) must be even no matter what n is => Sufficient
2) n<3 => n=2=> (m-11)(2-n) = 0 even => Sufficient
D is correct answer

(m-11)(2-n)
2m-mn-22+n

St.1- No information about value of n, not sufficient
St.2- only one prime number less than 3 is 2. So, n=2

even-even-even+even=even
Confirmed No, Sufficient

B

Given \((m - 11)(2 - n)\), and both m and n are prime, we have to check whether the value of this is odd.

Evaluating statement 1: \(m > 11\),

Since m is prime and greater than 11, m must be odd as 2 is the only even prime.
So for (m - 11) \(\implies\) (odd - odd) which is even.
So the multiplication with an even number is always even.
So this statement is sufficient.

Evaluating statement 2: \(n < 3\),

Since n is prime and less than 3, n must be 2 as it is the only prime < 3.
So for (2 - n) \(\implies\) (0) which is even.
So the multiplication with an even number is always even.
So this statement is sufficient.

Hence the answer is D IMO.

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

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

\((m-11)(2-n)\) can be rewritten as \(2(m-11) - n(m-11)\)

Now 2(m-11) is even. the expression evaluates to an odd number if n(m-11) is odd (since even - odd = odd) or else it will even (since even - even = even)
Also given, m and n are primes.

STATEMENT 1 : m>11
primes greater than 11 are ofcourse odd primes. Thus, n*(m-11) will be n*(even number) since odd - odd = even. So the expression will be even. So we can say it is NOT odd. thus statement one is sufficient

STATEMENT 2: n<3
prime < 3 is 2 , which is even prime. Thus, n*(m-11) will be 2*(m-11). So the expression will be even. So we can say it is NOT odd. thus statement two is sufficient

EACH STATEMENT ALONE IS SUFFICIENT. SO ANSWER IS D

If m and n are prime numbers, is (m - 11)(2 - n) an odd number ?
A product of two numbers is an even number if even one of the numbers is an even number.
(1) m > 11:-
There is only one even prime number, and that is "2". So all prime numbers greater than "11" must necessarily be odd.
Therefore it is established that "m" is an odd number. So (m-11) has to be an even number (sum or difference of two odd number is an even number always).

Hence (m-11)*(2-n) is an even number, and so statement (1) alone is sufficient.

(2) n < 3:-
The only prime number smaller than "3" is 2. Hence n=2 by given condition.

So (m-11)*(2-n) = 0, which is an even number.

Hence statement (2) alone is sufficient.

Correct ans is option D

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

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

Answer:
Case I: m>11 and m is a prime number
so m can be 13, 17, 19
so expression of (m-11) (2-n) will be (13-11)*(2-n) = 2*(2-n)
no matter what n , the expression becomes even
Also 2 is the only prime number so any prime number >11 will be odd and subtracting 11 from that odd makes it even
Hence this option looks good

Case II: n<3 and n is a prime number
which means n=2
the expression (m-11)(2-n) will be (m-11)*0 =0
and 0 is an even number
hence this option looks good

So D is the answer , each option looks good.

(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)

(m-11)(2-n) is odd when both m-11 and 2-n are odd. If one of the two factors is even, it's even

(1) m>11 and m is prime -> m must be odd, so m-11 must be even. Therefore, (m-11)(2-n) is even regardless of n -> Sufficient

(2) n<3 and n is prime -> only n=2, so 2-n=0 -> (m-11)(2-n)=0 (even) regardless of m -> Sufficient

Answer D

Given : m and n are prime numbers

To find : Whether (m - 11)(2 - n) is an odd number

Lets simplify the equation first -
(m - 11)(2 - n) = 2m -22 -mn +11n = (2m-22) - (mn-11n)
Here (2m-22) is Even
So we need to find if mn-11n is odd, Since Even - Odd = Odd
i.e, if n(m-11) = odd
this will be possible only when n is odd AND m-11 is odd
For m-11 to be odd , m has to be even.
Thus, the question becomes

Whether n is odd AND m is even?
If yes then only (m - 11)(2 - n) will be an odd number

(1) m > 11

All prime numbers >11 are odd
So m= odd

We wanted both n to be odd AND m to be even.
Now since m is odd, we can say that (m - 11)(2 - n) will not be an odd number .

Sufficient.


(2) n < 3

There is only 1 prime number less than 3.

2, which is even .
Thus, n is even.

We wanted both n to be odd AND m to be even.
Now since n is even, we can say that (m - 11)(2 - n) will not be an odd number .

Sufficient

Answer: D

Answer is each statement is enough

a) m>11, if m>11 and is a prime number then m is always odd. This mean that m-11 or odd-odd is even, hence the whole term (m-11)(2-n) will be even.

b) n<3; if n is less than 3 and is prime then the only value that n can take is 2. So, this makes the term 2-n as 0 and anything multiplied by 0 is 0. So the whole term will become even.

If m and n are prime numbers, is (m - 11)(2 - n) an odd number ?
Prime numbers are always positive integers
(1) m > 11
Since all prime number > 2 are odd, m>11 implies m is odd integer
hence, m-11 is an even integer. Therefore, (m-11)*(2-n) is Even number.
Sufficient
(2) n < 3
The only prime number less than 3 is 2. Hence, (2-n) = 0
Thus, (m-11)*(2-n) = 0 and we know that 0 is an Even integer.
Hence, statement 2 is sufficient.
Answer D

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

(1) m > 11
So m is prime and >11 and we also know that all primes except 2 are odd.
From this we can state that m is an odd prime be it 13 or even 17 etc.
Now odd - odd = even = 2k
Therefore m-11 = 2k and the parent equation becomes 2k(2 - n)
From this we can state that in any case (m-11)(2-n) will be even and NOT odd.
Hence (1) makes sense.
cancel out B,C,E

(2) n < 3
Now check statement 2.
we know n<3 and n is prime. So n should be 2.
substituting the value of n here (m - 11)(2 - n) we get (m-11)(2-2) = 0.
So we know that the value of (m - 11)(2 - n) is NOT odd.

Hence we can say both statements are self sufficient.
Thus IMO Option D

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

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

Let us do it.

1)(m - 11)(2 - n)
if m is equal to 17,19,23,.., this function (m - 11) is always equal to even number, and when i multiple any number to even number,result will be even number ->>>> example m=17 n=2 ->> 6*0=0 even number

Sufficient

2) n<3
there is only one prime number which is 2. Although i use different prime numbers for m, result will be always 0. Example, n=2 and m=13, 2*0=0-> even number

Sufficient

(1) m is a prime number and greater than 11. Because 2 is only even prime number, (m -11) is an even number. Therefore (m-11)(2-n) is an even number. Sufficient.
(2) n is a prime number and < 3. We have n = 1 or 2. So (2-n) = 1 or 0
when m = 2, (m - 11) is odd. We cannot determine whether the value of (m - 11) (2 - n) is odd or even.
(2) is insufficient.
Option A is the answer.