If m and n are positive integers such that m>n, what is t

[AAPL]

Veritas Prep

If m and n are 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 21 is 1.
2) The remainder when (m-n) is divided by 21 is 1.

OA C

[Brent@GMATPrepNow]

Veritas Prep

If m and n are 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 21 is 1.
2) The remainder when (m-n) is divided by 21 is 1.

OA C

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

Answer: C

[fskilnik@GMATH]

Veritas Prep

If m and n are 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 21 is 1.
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.)

$$\eqalign{
& m > n \ge 1\,\,\,{\rm{ints}}\,\,\,\left( * \right) \cr
& {m^{\rm{2}}} - {n^2} = 21K + R \cr
& K,R\,\,{\rm{ints}}\,\,,\,\,\,0 \le R \le 20 \cr
& ? = R \cr} $$

$$\eqalign{
& \left( 1 \right)\,\,m + n = 21J + 1\,\,,\,\,\,J\mathop \ge \limits^{\left( * \right)} 1\,\,\,{\mathop{\rm int}} \cr
& \left\{ \matrix{
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {21,1} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m^{\rm{2}}} - {n^2} = {21^2} - {1^2}\,\,\, \Rightarrow \,\,\,\,R = \,\,20 \hfill \cr
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {20,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m^{\rm{2}}} - {n^2} = {20^2} - {2^2} = \left( {20 - 2} \right)\left( {20 + 2} \right) = \left( {21 - 3} \right)\left( {21 + 1} \right) = 21\left( {21 + 1 - 3} \right) - 3\,\,\, \Rightarrow \,\,\,\,R = \,\,18\, \hfill \cr} \right. \cr} $$

$$\eqalign{
& \left( 2 \right)\,\,m - n = 21L + 1\,\,,\,\,\,L\mathop \ge \limits^{\left( * \right)} 0\,\,\,{\mathop{\rm int}} \, \cr
& \left\{ \matrix{
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {2,1} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m^{\rm{2}}} - {n^2} = 3\,\,\, \Rightarrow \,\,\,\,R = \,\,3 \hfill \cr
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {3,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m^{\rm{2}}} - {n^2} = 5\,\,\, \Rightarrow \,\,\,\,R = \,5\, \hfill \cr} \right. \cr} $$

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

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

Regards,
Fabio.