GMAT Club World Cup 2022 (DAY 7): If m and n are distinct prime number

From the question stem we know that m and n are distinct prime numbers. All prime numbers are odd except for 2.

Therefore if m and n is not 2, they are odd.

Question

Remainder { \(\frac{mn - (m+n) }{ 4}\) }

Statement 1

Both m! and n! are multiples of 4.

This means that both m and n are greater than 4, hence neither m nor n is 2. Therefore both m and n are odd.

Now, if m and n are both odd, mn - (m+n) is odd, hence the remainder can be either 1, 2 or 3.

Now we can represent a prime number as either (6n+1) or (6n-1)

m = (6x + 1) or (6x - 1)
n = (6y + 1) or (6y - 1)

In any of the above combination the remainder will always be 3.

Hence this statement is sufficient.

Statement 2

Neither \(m^n\) nor \(n^m\) is a multiple of 4.

This statement tells us, that neither m nor n is 2. Therefore both m and n are odd.

From our analysis at statement 1 we know that the remainder will always be 3.

Hence this statement is sufficient as well.

IMO D

S1:
"Both m! and n! are multiples of 4"
If m and n are distinct prime numbers and m! and n! are multiples of 4 then m and must be >=5 [m not equal n and m and are prime]
The smallest prime number possible is 5.
We know any prime greater than 3 can be written in the form 6k+1 OR 6k-1. [k=integer > 0] and [m and n both are >=5]

Consider below cases:

Case 1 m=6K1+1 and n=6K2+1
Thus m*n-(m+n)=36*K1*K2+6K1+6K2+1-6K1-6K2-2=36*K1*K2-1. -> This when divided by 4 leaves a reminder 4-1=3

Case 2 m=6K1-1 and n=6K2+1
Thus m*n-(m+n)=36*K1*K2-6K1-6K2+1+6K1+6K2-2=36*K1*K2-1. -> This when divided by 4 leaves a reminder 4-1=3

Case 3 m=6K1-1, n=6K2+1
Thus m*n-(m+n)=36*K1*K2+6K1-6K2-1+6K1+6K2=36*K1*K2+12K1 - 1 -> This when divided by 4 leaves a reminder 4-1=3

In each case reminder is 3. So statement is SUFFICIENT


S2:
"Neither m^n nor n^m is a multiple of 4."
What this statement indirectly tells us is that m and n cannot be 2, since 2 raised to higher powers will always be divided by 4.
As part of derivation of S1 we have seen if m and n are greater than 3 the given expression always gives a reminder of 3.

Now ALL prime numbers except 2 are ODD numbers. If we find the reminder of the expression divided by 4 using two odd numbers we can deduce the same conclusion for prime numbers as well.

Case
Let m=2p+1 and n=2q+1 [p and q are integers >0]
Thus m*n - (m+n)=(2p+1)*(2q+1)-2p-2q-2=4pq-1 -> This when divided by 4 leaves a reminder 4-1=3
This will be true for any prime which is not equal to 2

Thus this statement too is SUFFICIENT.

Ans D

Imo D

We have to find the remainder of [(m*n) - (m +n)]/4 =?

Statement 1: Both m! and n! are multiples of 4
Implies m>=4 and n >=4
Possible values of m = 5, 7, 11,13,17
Possible values of n = 5, 7, 11,13,17
[ m and n are distinct prime numbers]

Take m =5 and n= 7 then m*n = 35 and m + n= 12
So the remainder of (35-12)/4 = 3
Similarly test for multiple values such as -
m=5, n=11
n=5, m=13
The remainder in each case is coming to be 3.
Sufficient

Statement 2: Neither m^n nor n^m is a multiple of 4.
Implies that m and n cannot be 2.
Take m= 3 and n=5
Remainder (15-8)/4 =3
Take m= 3 and n=7
Remainder (21-10)/4 =3
Take m= 3 and n=11
Remainder (33-14)/4 =3
Sufficient

Answer: D

If m and n are distinct prime numbers, what is the remaindered when the positive difference between their product and their sum is divided by 4 ?

We need to find the remainder of |mn - (m+n)| / 4
Since, m and n are distinct prime numbers, they can be represented as
m = 6a+1 or 6a-1
n = 6b+1 or 6b-1

So, mn - (m+n) =
(6a+1)(6b+1) - 6a -1 - 6b - 1 = 36ab - 1 ....(1)
or
(6a-1)(6b-1) - 6a + 1 - 6b +1 = 36ab - 12a - 12b + 3 ........(2)

(1) Both m! and n! are multiples of 4.
=> m and n are both prime which are greater than 3.
i.e. m and n cannot be 2 or 3.
Now, for any prime greater than 2 or 3, if we substitute the values in expression (1) and (2), its clear that the remainder will be 3 when divided by 4.
Eg. for 5 and 7, (5*7 - 5 -7 )/4 = (35 - 12)/4 = 23/4 ... remainder is 3.
Sufficient.

(2) Neither m^n nor n^m is a multiple of 4.
=> m or n cannot be 2.
Again for any prime greater than 2, if we substitute the values in expression (1) and (2), its clear that the remainder will be 3 when divided by 4.
Eg. for 3 and 5, (3*5 - 3 - 5)/4 = (15-8)/4 = 7/4 ... remainder is 3.
Sufficient.

Given that m and n are distinct prime numbers.
From st.1 we have both m! and n! are multiples of 4. This suggests that m and n are not 2 or 3. i.e. they can be any distinct primes > 2 and 3.
For any primes considering the above, when positive difference between their product and their sum is divided by 4 we have: (|(m * n) – (m + n)|) / 4
In any case putting values of primes above we see that the remainder is always 3.
e.g. m = 5, n = 7 then remainder is: 3
if m = 13 , n = 19 then remainder is: 3
if m = 31, n = 37 then remainder is: 3
Hence this statement is sufficient.

From st.2 we have neither m^n nor n^m is a multiple of 4. This means neither of them is 2. In this case one of them can be 3.
In this case also we see that the remainder is always 3 for (|(m * n) – (m + n)|) / 4
e.g. if m = 3, n = 31 then remainder is: 3
if m = 3, n = 79 then remainder is: 3
Hence this statement is sufficient.

Hence answer choice (D)