If m and n are positive integers, is m + n divisible by 4 ?

Statement 1 : if m and n take 2 and 4 then yes
But if the take 6 and 4 then No. Insufficient.

Statement 2: if 2 and 6 then yes.
If 7 and 3 , No. Insufficient.
Combining we get a positive YES.

THUS IMO answer is C.
I hope this time I have not missed anything [FACE SAVOURING DELICIOUS FOOD]

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THUS IMO answer is C.



if u combine both
10 and 6 results in yes
and
10 and 14 results in Yes

so c is the answer

Thanks !!!

I missed..

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Statement 1
- Let's say m = (2*a*b*...), which a,b,... are prime numbers.
- Let's say n = (2*x*y*...), which x,y,... are prime numbers.
- \(\frac{(m+n)}{4}\) = \(\frac{(2*a*b*...) + (2*x*y*...)}{4}\) = \(\frac{2 ( (a*b*...) + (x*y*...))}{4}\)
- Here we can see that a,b,x,y can be anything prime numbers.
- #1 If (a*b) + (x*y) = EVEN, YES, \(\frac{m+n}{4}\) divisible by 4.
- #2 If (a*b) + (x*y) = ODD, NO, \(\frac{m+n}{4}\) cannot divisible by 4.
- Hence, INSUFFICIENT.

Statement 2
- Each of m and n has only one maximum one factor of 2 so it cannot divisible by 4.
- Try to plug simple number : m = 3 and n = 5, divisible by 4, YES.
- Try to plug another random number : m = 5 and n = 9, NOT divisible by 4, NO.
- Hence, INSUFFICIENT.

Both statement
- Go back to this equation :
- \(\frac{(m+n)}{4}\) = \(\frac{(2*a*b*...) + (2*x*y*...)}{4}\) = \(\frac{2 ( (a*b*...) + (x*y*...))}{4}\)
- If m & n divisible by 2 but not by 4, so a,b,x,y MUST BE ODD.
- (ODD*ODD) + (ODD*ODD) = ODD + ODD = EVEN.
- 2 * ANY even number MUST BE DIVISIBLE BY 4.
- Hence, SUFFICIENT.

If m and n are positive integers, is m + n divisible by 4 ?

(1) m and n are each divisible by 2.

(2) Neither m nor n is divisible by 4.

if m = 2 and n = 4 both are divisible by 2. Thus, satisfying the statement. However the sum is not divisible by 4.

if m = 2 and n = 2 both are divisible by 2 and their sum is divisible by 4.

Insufficient.

If n = 1 and m = 3 both are not divisible by 4. Thus, satisfying the statement. The sum is divisible by 4.

If n = 2 and m = 3 both are not divisible by 4. Thus, satisfying the statement. The sum is not divisible by 4.

Insufficient.

Now we need numbers that satisfy both statements, the easiest ones I could think of is 2 and 6 both are divisible by 2 and both are not divisible by 4.

6/2 = 3 and 2/2 = 1
6/4 = 1.5 and 2/4 = 0.5

6 + 2 = 8/4 = 2

Combined is sufficient.

C is the answer choice

Consider 2,4,6,8,10

Statement 1: 2+4 not div by 4. 2+6 is divisible by 4. Not sufficient.

Statement 2: Clearly not sufficient. m and n can be any positive numbers.

Combined: If you remove the multiples of 4, then the sum is always a multiple of 4. (2+6,6+6,2+10)
Hence C

Combining 1 & 2,

m and n are divisible by 2 but not by 4, which means both are ODD multiples of 2.
Odd multiples of 2 can be written in the form :
m = 2(2k+1)
n = 2(2p+1)
=> m+n = 4(k+p)+ 4, which is clearly divisible by 4

Statement 1:
Case 1: m=2 and n=2, with the result that m+n=4
In this case, m+n is divisible by 4, so the answer to the question stem is YES.
Case 2: m=2 and n=4, with the result that m+n=6
In this case, m+n is not divisible by 4, so the answer to the question stem is NO.
Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.

Statement 2:
Case 1: m=2 and n=2, with the result that m+n=4
In this case, m+n is divisible by 4, so the answer to the question stem is YES.
Case 3: m=1 and n=2, with the result that m+n=3
In this case, m+n is not divisible by 4, so the answer to the question stem is NO.
Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.

Statements combined:
Since m is divisible by 2 but not by 4, m = 2a, where a is ODD.
Since n is divisible by 2 but not by 4, n = 2b, where b is ODD.
Thus:
m+n = 2a + 2b = 2(a+b) = 2(ODD + ODD) = 2(EVEN) = multiple of 4
The answer to the question stem is YES.
SUFFICIENT.

If m and n are positive integers, is m + n divisible by 4 ?

(1) m and n are each divisible by 2.

Let's choose two numbers: m = 2, n = 4
(m+n) is not divisible by 4

m = 2 and n = 2
(m+n) is divisible by 4

Thus, not sufficient

(2) Neither m nor n is divisible by 4

Let's take two numbers: m = 7, n = 5
(m+n) is divisible by 4

m = 2, n = 7
(m+n) is not divisible by 4

Thus not sufficient.

Lets combine (1) and (2),

m has to be divisible by 2 but not by 4. Similarly, n has to be divisible by 2 but not by 4

m = n = 2, thus (m+n) is divisible by 4

m = 6, n = 2, still (m+n) is divisible by 4.

Thus sufficient.


Another way to look at this combined statement is that, for any number which is divisible by 2 and 4, if we substract 2 from that number, then that number becomes divisible by only 2. Thus take any such number divisible by 2 and 4, say 16. Now, 16-2 = 14 is divisible by only 2. Thus all we are doing is taking a number x divisible by 2 and 4 both and taking a number y divisible by 2 and 4 both and eventually substracting 2 from x and 2 from y resulting in (x+y-4). This resulting number is always going to be divisible by 4

If m and n are positive integers, is m + n divisible by 4 ?

(1) m and n are each divisible by 2.
if m = n = 2; m + n is divisible by 4
if m = 2 and n = 4; m + n is not divisible by 4

INSUFFICIENT.

(2) Neither m nor n is divisible by 4.
if m = 6 and n = 1; m + n is not divisible by 4
if m = 7 and n = 1; m + n is divisible by 4

INSUFFICIENT.

(1+2) Neither m nor n is divisible by 4 AND m and n are divisible by 2

if m = 2 and n = 6; m + n is divisible by 4
if m = 10 and n = 6; m + n is divisible by 4

SUFFICIENT.

Answer is C.

Solution:

We need to determine whether m + n is divisible by 4, given that m and n are positive integers.

Statement One Alone:

Statement one alone is not sufficient. For example, if m = 2 and n = 2, then m + n = 4 is divisible by 4. However, if m = 2 and n = 4, then m + n = 6 is not divisible by 4.

Statement Two Alone:

Statement two alone is not sufficient. For example, if m = 2 and n = 2, then m + n = 4 is divisible by 4. However, if m = 1 and n = 2, then m + n = 3 is not divisible by 4.

Statements One and Two Together:

Since m is divisible by 2 but not by 4, m can be expressed as 2 times an odd number, i.e,. 2(2k + 1) for some integer k. Likewise, since n is divisible by 2 but not by 4, n can be expressed as 2 times an odd number, i.e,. 2(2j + 1) for some integer j. Therefore,

m + n = 2(2k + 1) + 2(2j + 1) = 2(2k + 1 + 2j + 1) = 2(2k + 2j + 2) = 2[2(k + j + 1)] = 4(k + j + 1)

We see that m + n is divisible by 4 since it’s a multiple of 4. Both statements together are sufficient.

Answer: C

KarishmaB

I am a bit confused on proving that the statements combined can be useful. How do we know to define a and b as both odd numbers? I would be so appreciative for your insights, and if you may have another way to efficiently arrive at the answer C.

Hi woohoo921
Thanks for your query.


Let me give you another method to solve this problem in which I will not use plugging in as the main approach. Instead, I will show you the methodical approach to help you to arrive at the correct answer choice.


For this, first, let us analyze the question stem properly.


QUESTION STEM ANALYSIS:
From the question stem, we only got that m and n are POSITIVE INTEGERS. And for these m and n, we need to find whether (m + n) is divisible by 4.

Let’s just first translate the entire question stem from English to Math. It has two parts- given and to find:
- Given: m and n belong to the set of positive integers {1, 2, 3, 4, …}
- To find: Whether (m + n) = 4k, for some positive integer k.

Note: You must always translate as you go. This will help you focus on the main workable elements of a question – the juice you extract from all its wordy sentences.

As this is all we can get from the question stem, let’s move to statement 1!


STATEMENT 1 ANALYSIS: “m and n are each divisible by 2.”
This implies that m and n can be represented as 2a and 2b, respectively, for positive integers ‘a’ and ‘b’. (Observe how we translated English to Math here as well, as our first task)


So, let’s check what we can now say about (m + n).

• m + n = 2a + 2b = 2(a + b).
• Since we finally need to check whether m + n = 4k or not, we must further analyze the nature of (a + b). There are two possibilities:
  
  1. If (a + b) is even: Then, (a + b) = 2k, (k is a positive integer), and hence, (m + n) = 2(2k) = 4k.
     
     • So, here, the answer to the main question asked is YES!
  2. If (a + b) is odd: Then, (a + b) = 2k + 1, and hence, (m + n) = 2(2k + 1) = 4k + 2.
     
     • Since 4k + 2 is not divisible by 4, the answer to our main question is NO!

As we are not getting a unique answer from statement 1 alone, it is INSUFFICIENT.
Next, let’s move ahead and analyze statement 2.


STATEMENT 2 ANALYSIS: “Neither m nor n is divisible by 4.”
In simple terms, m ≠ 4p and n ≠ 4q for any integers p and q, respectively.

So, m can take any form out of 4p + 1, 4p + 2, 4p + 3 but not 4p + 4 (this will again mean 4(p + 1) and hence, a multiple of 4). Similarly, n can take any form out of 4q + 1, 4q + 2, 4q + 3 but not 4q + 4.


Now, there can be many possible combinations of m and n that will give us different expressions for (m + n). Below are just two of those possibilities:

• If m = 4p + 1 and n = 4q + 1:
  
  • m + n = 4(p + q) + 2 = 4(some integer) + 2. -> NOT a multiple of 4
  • The answer to the main question asked is NO!
• If m = 4p + 2 and n = 4q + 2:
  
  • m + n = 4(p + q) + 4 = 4(p + q + 1) = 4(some integer). -> a multiple of 4
  • The answer to the main question asked is YES!

Again, we did not get a unique answer from statement 2 alone, making it INSUFFICIENT.

Since none of the statements alone comes out to be sufficient, let’s move ahead and analyze combined statements 1 and 2.


STATEMENT 1 AND 2 TOGETHER:
From statement 1, we know that m = 2a and n = 2b, for integers a and b. And, from statement 2, we know that m ≠ 4p and n ≠ 4q, for any integers p and q.

So, m and n have exactly one 2 as their factor. (If m and n had two 2s, they would also have a 4 which is not allowed per statement 2). Thus, we want such values of a and b, that do not contribute another 2 to m and n.

Now, a and b can contribute another 2 to m and n only if they are even. Since we do not want this to happen, ‘a’ and ‘b’ are NOT EVEN. That is, a and b are both ODD.


Finally, if a and b are odd, we can express them as a = 2p + 1 and b = 2r + 1, for some integers p and r.
So, m + n = 2a + 2b can be simplified as:

• = 2(2p + 1) + 2(2r + 1) = 4p + 2 + 4r + 2
• = 4p + 4r + 4 = 4(p + r + 1)
• = 4k (divisible by 4)

And that’s it! The answer to our main question is a sure YES!
Thus, statements 1 and 2 combined are sufficient to answer our question.
Hence, option C is the correct answer.


Hope this helps!


Best,
Aditi Gupta
Quant expert, e-GMAT.

Rule:
(EVEN)(EVEN) = MULTIPLE OF 4

Statement 1 implies that m = 2a and n = 2b, since m and n are both divisible by 2.
In accordance with the rule above:
If \(a\) is even, then m = 2a = (EVEN)(EVEN) = MULTIPLE OF 4.
Since Statement 2 requires that m not be divisible by 4, a\(\) cannot be even --> \(a\) is ODD.
Similarly:
If \(b\) is even, then n = 2b = (EVEN)(EVEN) = MULTIPLE OF 4.
Since Statement 2 requires that n not be divisible by 4, \(b\) cannot be even --> \(b\) is ODD.

Bunuel and other experts, How can we assume in this question that m' and n' are distinct positive integers if not explicitly mentioned ?