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Re: If m and n are positive integers such that m > n, what is the remainde [#permalink]
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m,n +ve I. Find remainder when (m+n)(m-n)/21 :

1) (m+n) = 7x+1 => clearly not sufficient since we dont know about m-n
2) (m-n) = 3y+1 => Similarly NS.

1) + 2) => (m+n)(m-n) = 21xy + 7x + 3y + 1
=> (m^2 - n^2) /21 = xy + (7x+3y+1)/21

Thus remainder depends on the term (7x+3y+1)/21 => No definite value. Thus NS .

Ans E.
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Re: If m and n are positive integers such that m > n, what is the remainde [#permalink]
My solution was as follows:
To me it was fairly obvious that both options were insufficient separately, for instance, giving a quick glance at the first option
\(m + n = 7*k + 1\)
\((7*k+1)*(m-n) = 21*l + X\)
There is simply no way you can figure out what X is from this equation. Respectively, 2nd option is almost the same.
Lets combine them now:
\(m + n = 7*k + 1\)
\(m - n = 3*f + 1\)
\(m^2 - n^2 = 21*f*k + (7*k+3*f + 1) = 21*b + X\)
The second equation tells us that \(X = 7*k+3*f+1\) and \(f*k = b\)
Since we are not really limited with number choices, all we have to do is choose 2 different values of b that would yield us relevant (m;n) values.
1: \(b = 1\) => \(f = 1, g = 1\) => \(m = 6, n = 2, X = 7*1 + 3*1 + 1 = 11\)
2: \(b = 3\) => \(f = 3, g = 1\) => \(m = 13, n = 9, X = 7*3 + 3*1 + 1 = 25\)
Thus, even with 2 options combined we can't define X, thus option E is the answer
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Re: If m and n are positive integers such that m > n, what is the remainde [#permalink]
Expert Reply
Hi All,

This DS question can be approached a number of different ways. Here's an approach that focuses on TESTing VALUES.

We're told that M and N are POSITIVE INTEGERS and that M > N. We're asked for the remainder when M^2 - N^2 is divided by 21.

Fact 1: The remainder when (m + n) is divided by 7 is 1.

IF....
M= 6
N = 2
8/7 = 1r1
The answer to the question is (36-4)/21 = 1r11 so the answer is 11.

IF....
M = 14
N = 1
15/7 = 2r1
The answer to the question is (196-1)/21 = 9r6 so the answer is 6
Fact 1 is INSUFFICIENT

(2) The remainder when (m – n) is divided by 3 is 1.

IF....
M= 6
N = 2
4/3 = 1r1
The answer to the question is (36-4)/21 = 1r11 so the answer is 11.

IF....
M = 14
N = 1
13/3 = 4r1
The answer to the question is (196-1)/21 = 9r6 so the answer is 6
Fact 2 is INSUFFICIENT

Combined, we have a set of values that 'fits' both Facts and provide different answers:
IF....
M= 6
N = 2
4/3 = 1r1
The answer to the question is (36-4)/21 = 1r11 so the answer is 11.

IF....
M = 14
N = 1
13/3 = 4r1
The answer to the question is (196-1)/21 = 9r6 so the answer is 6

Combined, INSUFFICIENT

Final Answer:

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Re: If m and n are positive integers such that m > n, what is the remainde [#permalink]
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Expert Reply
Bunuel wrote:
If m and n are positive integers such that m > n, what is the remainder when m^2 – n^2 is divided by 21?

(1) The remainder when (m + n) is divided by 7 is 1.
(2) The remainder when (m – n) is divided by 3 is 1.


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

First let’s give you the incorrect solution provided by many.

Question: What is the remainder when (m^2 – n^2) is divided by 21?

Statement 1: The remainder when (m + n) is divided by 7 is 1.

(m + n) = 7a + 1

Statement 2: The remainder when (m – n) is divided by 3 is 1.

(m – n) = 3b + 1

Therefore, remainder of product (m^2 – n^2) = (m + n)*(m – n) = (7a + 1)(3b + 1) when it is divided by 21 is 1.

Answer (C)

This would have been correct had the statements been:

Statement 1: The remainder when (m + n) is divided by 21 is 1.

Statement 2: The remainder when (m – n) is divided by 21 is 1.

Statement 1: (m + n) = 21a + 1

Statement 2: (m – n) = 21b + 1

(m^2 – n^2) = (m + n)*(m – n) = (21a + 1)*(21b + 1) = 21*21ab + 21a + 21b + 1

Here, every term is divisible by 21 except the last term 1. So when we divide (m^2 – n^2) by 21, the remainder will be 1.

But let’s go back to our original question. If you solved it the way given above and got the answer as (C), you are not the only one who jumped the gun. Many people end up doing just that. But here is the correct solution:

The statements given are:

Statement 1: The remainder when (m + n) is divided by 7 is 1.

(m + n) = 7a + 1

Statement 2: The remainder when (m – n) is divided by 3 is 1.

(m – n) = 3b + 1

This gives us (m^2 – n^2) = (m + n)*(m – n) = (7a + 1)(3b + 1) = 21ab + 7a + 3b + 1

Here only the first term is divisible by 21. We have no clue about the other terms. We cannot say that 7a is divisible by 21. It may or may not be depending on the value of a. Similarly, 3b may or may not be divisible by 21 depending on the value of b. So how can we say here that the remainder must be 1? We cannot. We do not know what the remainder will be in this case even with both statements together.

Say, if a = 1 and b = 1,

m^2 – n^2 = 21*1*1 + 7*1 + 3*1 + 1 = 21 + 11

The remainder when you divide m^2 – n^2 by 21 will be 11.

Say, if a = 2 and b = 1,

m^2 – n^2 = 21*2*1 + 7*2 + 3*1 + 1 = 21*2 + 18

The remainder when you divide m^2 – n^2 by 21 will be 18.

Hence, both statements together are not sufficient to answer the question.

Answer (E)
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If m and n are positive integers such that m > n, what is the remainde [#permalink]
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Bunuel wrote:
If m and n are positive integers such that m > n, what is the remainder when m^2 – n^2 is divided by 21?

(1) The remainder when (m + n) is divided by 7 is 1.
(2) The remainder when (m – n) is divided by 3 is 1.­

­Statement 1: m+n = 7a+1 = 1, 8, 15...
Statement 2: m-n = 3b+1 = 1, 4, 7...


Case 1: m+n=8 and m-n=4, with the result that m=6, n=2 and \(m^2-n^2 = 36-4 = 32\)
When we divide \(m^2-n^2\) by 21, we get:
\(\frac{32}{21} =\) 1 R11

Case 2: m+n=15 and m-n=1, with the result that m=8, n=7 and \(m^2-n^2 = 64-49 = 15\)
When we divide \(m^2-n^2\) by 21, we get:
\(\frac{15}{21} =\) 0 R15

Since the remainder can be different values, the two statements combined are INSUFFICIENT.

­
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If m and n are positive integers such that m > n, what is the remainde [#permalink]
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