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

IMO answer D

n,m = prime numbers => can be {2,3,5, 7,11,13,17,....etc}
n, m distincts

If |nm-n-m|=4k +x, x=?

1.=> n>=4, m>=4
Choice 1 {n, m}
{7,5}
|35-12|= 23=4*5 +3 => x=3

Choice 2
{5,11}
|55-16| =39 =4*9 +3 => x=3
Therefore, 1 is sufficient

2. n^m and m^n are not divisible by 4
Let's find some couple of n, m that satisfy that clause.
{3,5} => 3^5 =243 OK
5^3 = 125 ok

Then |nm - n-m| =7 =4 +3
x=3

2nd couple {7,3}
|21-10|=11 =4*2 +3
X=3

2 is sufficient.
Each alone sufficient => D.

Posted from my mobile device

What do we need to find?
Remainder when the positive difference between their product and their sum is divided by 4

Statement 1:
We can conclude that both m and n are > 4. Let's take the m=7, n=11. Product = 77, sum = 18. Positive Difference = 59. Remainder = 3.
The same thing happens if you take m = 5 or 11 and n = 7 or 13 and so on. The remainder is always 3.

Statement 2:
We can conclude that both m and n are odd and thus neither of them are 2.

Let m = 3, n = 11
Product = 33
Sum = 14
Difference = 19.
It's the same for all combinations. The remainder is always 3

Hence each statement alone is sufficient. Option D is the answer.

(1) Both m! and n! are multiples of 4: Tale 4 common, it will be 4{() -()} / 4. So remaindered will be 0. Sufficient
(2) Neither m^n nor n^m is a multiple of 4: Not sufficient, as for some value of m,n it gives 0 remainder and for some, other. Hence not sufficient.

Option A is the correct answer

Hope this helps!

Posted from my mobile device

Asked: 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 ?

What is the remainder when |mn-(m+n)| is divided by 4?

(1) Both m! and n! are multiples of 4.
Both m & n are greater than 3. Any prime number greater than 2 is odd.
Let m = 2k+1; n=2l + 1
Since mn > (m+n);
|mn - (m+n) = (2k+1)(2l+1) - (2k+1+2l+1) = 4kl + 2(k+l) + 1 - 2(k+l) -2
Remainder when 4kl + 2(k+l) + 1 - 2(k+l) -2 is divided by 4 = 4 -1 = 3
SUFFICIENT

(2) Neither m^n nor n^m is a multiple of 4.
Both prime numbers are odd and greater than 2.
Let m = 2k+1; n=2l + 1
Since mn > (m+n);
|mn - (m+n)| = (2k+1)(2l+1) - (2k+1+2l+1) = 4kl + 2(k+l) + 1 - 2(k+l) -2 ; k+l is even; 2(k+l) is divisible by 4.
Remainder when 4kl + 2(k+l) + 1 - 2(k+l) -2 is divided by 4 = 4 -1 = 3
SUFFICIENT

IMO D

Correct answer : Choice D

Statement 1 says : Both m! and n! are multiples of 4.

Since the question stem says distinct prime numbers and both have multiples of 4 in their factorials, it has to be prime numbers after or equal to 5.
Taking any prime number equal to after 5 and doing the operation given in the question stem always gives the remainder = 3. Hence statement 1 is sufficient .

Statement 2 says : Neither m^n nor n^m is a multiple of 4.
Given that they are prime numbers, the above statement is possible only if the prime numbers are above or equal to 3.
Taking any prime number equal to after 3 and doing the operation given in the question stem always gives the remainder = 3. Hence statement 2 is sufficient .

hence from the above , we can coclude that , Choice D is the correct answer

IMO , Option D is correct.

Statement 1 only. shows that the distinct prime numbers are not 2 or 3. The smallest prime number is at least 5.
The reason is that 2! = 2 (Not a multiple of 4)
and 3! = 6 (Not a multiple of 4)
The product of the two distinct integers must be odd. The sum of the integers will be even.
Try a few two distinct prime integers greater than 2 and 3.

Try 5 and 7. Product is 35. Sum is 12. Product -Sum =35-12 = 23. 23/4 gives Remainder 3.
Try 7 and 11. Product is 77. Sum is 18. Product -Sum =77-18 = 59. 59/4 gives Remainder 3.
Try 5 and 13. Product is 65. Sum is 18. Product -Sum =65-18 = 47. 47/4 gives Remainder 3.
Try 11 and 13. product is 143. Sum is 24. Product -Sum =143-24 = 119. 119/4 gives Remainder 3.
I can conclude Statement I alone is sufficient because I can see a pattern.

Statement 2 only shows that none of the distinct prime numbers is 2.

By testing different distinct prime numbers excluding 2, will give a remainder of 3 with the conditions in the question.
I can conclude Statement II alone is sufficient because I can see a pattern.

Therefore option D is the answer IMO.

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

(1) Both m! and n! are multiples of 4.

Sufficient

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

Sufficient

Ans - D

Solution attached

Posted from my mobile device

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 ?

(1) Both m! and n! are multiples of 4.
(2) Neither m^n nor n^m is a multiple of 4.


statement 1 - m and n should be 5, 7,......; remainder always be 3

sufficient

statement 2 - value of m and n should not be 2, which is a prime number
so, m and n should be 3, 5, 7,......; remainder always be 3

sufficient

answer 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

[quote="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 ?

(1) Both m! and n! are multiples of 4.
(2) Neither m^n nor n^m is a multiple of 4.

m and n are distinct prime numbers
Statement 1)Both m! and n! are multiples of 4 that means m, n >5 and neither of them can be 2.
As all prime numbers greater than 2 are odd thus we can consider m and n as 2p1+1 and 2p2 +1
Thus product of 2p1+1 and 2p2+1 =4p1p2+2p1+2p2+1
and sum of 2p1+1 and 2p2+1=2p1+2p2+2
Thus the difference =4p1p2-1=>As 4p1p2 is divisible by 4 the remainder is always -1 which is 3
thus 1 is sufficient
2)Neither m^n nor n^m is a multiple of 4.Thus we cannot consider 2 as prime number in this case.
Again we can consider the same example for odd prime numbers
we can consider m and n as 2p1+1 and 2p2 +1
Thus product of 2p1+1 and 2p2+1 =4p1p2+2p1+2p2+1
and sum of 2p1+1 and 2p2+1=2p1+2p2+2
Thus the difference =4p1p2-1=>As 4p1p2 is divisible by 4 the remainder is always -1 which is 3
Thus 2 is sufficient.
answer D

Given M and N are distinct primes. What is the remainder R when {m*n - (m+n)}/4 ?

I) m! and n! are multiples of 4 implies both primes are greater than 4
Consider, 7 & 5 we have R = 2
If we consider 11 & 7, R = 1

I not sufficient since two different R

II) m^n or n^m is a multiple of 4
Both sets of numbers (7,5) and (11,7) are still applicable and gives us two different remainders

II is not sufficient

Combining I and II - 7,5 and 11,7 are still applicable and hence the given statements are not enough to answer

Choice E is the correct answer

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

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
that is Remainder of (mn-(m+n))/4 ..m*n product of 2 odd numbers = Odd and m+n = Even, Odd-Even = Odd

(1) Both m! and n! are multiples of 4
since m and n are distinct and m! and n1 are multiples of 4, both m and n are prime numbers greater than 3
When we find remainder of (mn-(m+n))/4, we will always get remainder as 3 , hence sufficient

(2) Neither m^n nor n^m is a multiple of 4
this implies m and n are greater than 2, as if m=2, n=5 then 2^5 is multiple of 4.
hence when we find remainder of (mn-(m+n))/4, we will always get remainder as 3 , hence 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 ?

m & n can be either 2 (only even prime) or other prime odd number.
If m & n are both odd prime let m = 2k+1 and n = 2l+1

To find remainder of (mn - m - n)/4
=> [(2k+1)(2l+1) - (2k+1) - (2l+1) ] / 4
= [4kl + 2k + 2l + 2 - 2k - 1 - 2l -1] / 4
= [4kl - 1] /4
The remainder of this equation would be '3' always.
Hence, with the given statements we need to know whether m & n are both odd prime?

(1) Both m! and n! are multiples of 4.
=> This implies that both m & n are greater than 4. Hence, statement 1 alone is sufficient.

(2) Neither m^n nor n^m is a multiple of 4.
=> Implies neither m nor n is equal to '2' i.e., both m & n are odd primes. Hence, statement 2 alone is sufficient.

Answer: D