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# If m and n are positive integers such that m > n, what is the Re.

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

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17 Oct 2016, 12:14
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If m and n are positive integers such that m > n, what is the remainder when m^2 – n^2 is divided by 21?

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.

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

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17 Oct 2016, 12:30
stonecold 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?

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.

FROM STATEMENT - I ( INSUFFICIENT)

We do not have any relationship between the value of m & n as such it won't be possible for us to find the remainder of $$\frac{(m^2 – n^2)}{21}$$, since m & n can take any values...

FROM STATEMENT - I ( INSUFFICIENT)

We do not have any relationship between the value of m & n as such it won't be possible for us to find the remainder of $$\frac{(m^2 – n^2)}{21}$$, since m & n can take any values...

COMBINE STATEMENT I & II (SUFFICIENT)

$$\frac{(m^2 – n^2)}{21} = \frac{( m+n )( m-n )}{21}$$ = $$Remainder \ 1$$

Hence BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked, answer will be (C)

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

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29 Mar 2018, 01:34
stonecold 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?

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.

Bunuel, pushpitkc : Please bring some light on this question using number plugging technique.

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

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02 Jan 2019, 11:11
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stonecold wrote:
If m and n are positive integers such that m > n, what is the remainder when m² – n² is divided by 21?

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

Given: m and n are positive integers such that m > n

Target question: What is the remainder when m² – n² is divided by 21?

Statement 1: The remainder when (m + n) is divided by 21 is 1
There are several values of m and n that satisfy statement 1. Here are two:
Case a: m = 12 and n = 10. This means m + n = 12 + 10 = 22, and 22 divided by 21 leaves remainder 1. In this case, m² – n² = 12² – 10² = 144 - 100 = 44.
When we divide 44 by 21, we get 2 with remainder 2. So, the answer to the target question is when m² – n² is divided by 21, the remainder is 2
Case b: m = 13 and n = 9. This means m + n = 13 + 9 = 22, and 22 divided by 21 leaves remainder 1. In this case, m² – n² = 13² – 9² = 169 - 81 = 88.
When we divide 88 by 21, we get 4 with remainder 4. So, the answer to the target question is when m² – n² is divided by 21, the remainder is 4
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: The remainder when (m – n) is divided by 21 is 1
There are several values of m and n that satisfy statement 2. Here are two:
Case a: m = 5 and n = 4. This means m - n = 5 - 4 = 1, and 1 divided by 21 leaves remainder 1. In this case, m² – n² = 5² – 4² = 25 - 16 = 9.
When we divide 9 by 21, we get 0 with remainder 9. So, the answer to the target question is when m² – n² is divided by 21, the remainder is 9
Case b: m = 4 and n = 3. This means m - n = 4 - 3 = 1, and 1 divided by 21 leaves remainder 1. In this case, m² – n² = 4² – 3² = 16 - 9 = 7.
When we divide 7 by 21, we get 0 with remainder 7. So, the answer to the target question is when m² – n² is divided by 21, the remainder is 7
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that the remainder is 1 when (m + n) is divided by 21
In other words, m+n is 1 greater than some multiple of 21.
So, we can write: m+n = 21k + 1 (for some integer k)

Statement 2 tells us that the remainder is 1 when (m - n) is divided by 21
In other words, m-n is 1 greater than some multiple of 21.
So, we can write: m-n = 21j + 1 (for some integer j)

Now recognize that we can factor m² – n²
We get: m² – n² = (m + n)(m - n)
= (21k + 1)(21j + 1)
= 21²mn + 21k + 21j + 1
= 21(21mn + k + j) + 1
Since 21(21mn + k + j) is definitely a multiple of 21, we can conclude that 21(21mn + k + j) + 1 is 1 greater than some multiple of 21.
In other words, m² – n² is 1 greater than some multiple of 21.
So, when m² – n² is divided by 21, the remainder is 1
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

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

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03 Jan 2019, 04:13
stonecold 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?

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.

Important: the solution presented below assumes a reasonable quantitative maturity.
(If you do not have that, I recommend the excellent step-by-step presentation offered by Brent above.)

$$m > n \ge 1\,\,\,{\rm{ints}}\,\,\,\left( * \right)$$

$${m^{\rm{2}}} - {n^2} = 21K + R$$

$$K,R\,\,{\rm{ints}}\,\,,\,\,\,0 \le R \le 20$$

$$? = R$$

$$\left( 1 \right)\,\,m + n = 21J + 1\,\,,\,\,\,J\mathop \ge \limits^{\left( * \right)} 1\,\,\,{\mathop{\rm int}}$$

$$\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {21,1} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m^{\rm{2}}} - {n^2} = {21^2} - {1^2}\,\,\, \Rightarrow \,\,\,\,R = \,\,20 \hfill \cr \,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {20,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m^{\rm{2}}} - {n^2} = {20^2} - {2^2} = \left( {20 - 2} \right)\left( {20 + 2} \right) = \left( {21 - 3} \right)\left( {21 + 1} \right) = 21\left( {21 + 1 - 3} \right) - 3\,\,\, \Rightarrow \,\,\,\,R = \,\,18\, \hfill \cr} \right.$$

$$\left( 2 \right)\,\,m - n = 21L + 1\,\,,\,\,\,L\mathop \ge \limits^{\left( * \right)} 0\,\,\,{\mathop{\rm int}} \,$$

$$\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {2,1} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m^{\rm{2}}} - {n^2} = 3\,\,\, \Rightarrow \,\,\,\,R = \,\,3 \hfill \cr \,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {3,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m^{\rm{2}}} - {n^2} = 5\,\,\, \Rightarrow \,\,\,\,R = \,5\, \hfill \cr} \right.$$

$$\left( {1 + 2} \right)\,\,\,\,\left( {21J + 1} \right)\left( {21L + 1} \right) = 21\left( {21JL + J + L} \right) + 1\,\,\, \Rightarrow \,\,\,\,R = \,\,1\,\,\,\,$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: If m and n are positive integers such that m > n, what is the Re. &nbs [#permalink] 03 Jan 2019, 04:13
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