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Could someone explain how n = 7p + 5 was derived?
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Could someone explain how n = 7p + 5 was derived?

There is a link provided in the solution: ...check HERE to know to to derive general formula from these two...
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(1) n = 13q+3 (3,16,29,...)

if we try to divide each possible value of n by 2, we find R is not consistent (i.e. R = 1, R=0, R=1...)

NOT SUFFICIENT

(2) n+2 =7#

say n+2=7 --> n=5 --> 5/2 --> R = 1
say n+2=14 --> n=12 --> 12/2 --> R = 0

inconsistent remainders, thus NOT SUFFICIENT

(1) + (2)
13q+5=7#

13 is a prime number. There is no way we would be able to obtain 13 by subtracting 5 from a multiple of 7.

Thus, insufficient.

E
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What is the remainder when the positive integer n is divided by 2?

(1) When n is divided by 13, the remainder is 3
(2) n + 2 is a multiple of 7

We need to determine the remainder when the positive integer n is divided by 2. Keep in mind that when a positive integer is divided by 2, the remainder is either 0 (if the number is even) or 1 (if the number is odd).

Statement One Alone:

When n is divided by 13, the remainder is 3.

There are many possible values for n. For example, n could be 16 or n could be 29. If n = 16, the remainder is 0 when n is divided by 2. However, if n = 29, the remainder is 1 when n is divided by 2. Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

n + 2 is a multiple of 7.

There are many possible values for n. For example, n could be 5 or n could be 12. If n = 5, the remainder is 1 when n is divided by 2. However, if n = 12, the remainder is 0 when n is divided by 2. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Even with the two statements, there are still many possible values for n. For example, n could be 68 or n could be 159. If n = 68, the remainder is 0 when n is divided by 2. However, if n = 159, the remainder is 1 when n is divided by 2. The two statements together are still not sufficient to answer the question.

Answer: E
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What is the remainder when the positive integer n is divided by 2?

The remainder upon division n by 2 can be 0 (when n is even) or 1 (when n is odd). So, the question basically asks whether n is even or odd.

(1) When n is divided by 13, the remainder is 3 --> n = 13q + 3 --> n can be: 3, 16, 29, 42, 55, 68, ... So, n can be even as well as odd. Not sufficient.

(2) n + 2 is a multiple of 7 --> n is 2 less than a multiple of 7, or, which is the same, n is 5 more than a multiple of 7 --> n = 7p + 5: 5, 12, 19, 26, 33, 40, 47, 54, 61, 68, ... So, n can be even as well as odd. Not sufficient.

(1)+(2) From n = 13q + 3 and n = 7p + 5, we can get that n = 91k + 68 (check HERE to know to to derive general formula from these two), so n can be 68, 159, ... So, again, n can be even as well as odd. Not sufficient.

Answer: E.

Hope it's clear.

Hi Bunuel,

Could you please advise if we can just add one equation to another? In that case we would have n=20k+1. By picking the numbers we see that it can either be even OR odd, so we get E as the answer.
This approach is presented by MGM for tackling extra remainders problems.

Thanks.
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Bunuel
What is the remainder when the positive integer n is divided by 2?

The remainder upon division n by 2 can be 0 (when n is even) or 1 (when n is odd). So, the question basically asks whether n is even or odd.

(1) When n is divided by 13, the remainder is 3 --> n = 13q + 3 --> n can be: 3, 16, 29, 42, 55, 68, ... So, n can be even as well as odd. Not sufficient.

(2) n + 2 is a multiple of 7 --> n is 2 less than a multiple of 7, or, which is the same, n is 5 more than a multiple of 7 --> n = 7p + 5: 5, 12, 19, 26, 33, 40, 47, 54, 61, 68, ... So, n can be even as well as odd. Not sufficient.

(1)+(2) From n = 13q + 3 and n = 7p + 5, we can get that n = 91k + 68 (check HERE to know to to derive general formula from these two), so n can be 68, 159, ... So, again, n can be even as well as odd. Not sufficient.

Answer: E.

Hope it's clear.

Hi Bunuel,

Could you please advise if we can just add one equation to another? In that case we would have n=20k+1. By picking the numbers we see that it can either be even OR odd, so we get E as the answer.
This approach is presented by MGM for tackling extra remainders problems.

Thanks.

No, this would be incorrect.

You cannot add n = 13q + 3 and n = 7p + 5 to get n = 20k + 1. Notice that the quotients there are different: q and p. The way you can get the general formula is given in the solution.
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TARGET730
What is the remainder when the positive integer n is divided by 2?

(1) When n is divided by 13, the remainder is 3
(2) n + 2 is a multiple of 7

We need to determine the remainder when n is divided by 2.

Statement One Alone:

When n is divided by 13, the remainder is 3.

Statement one alone is not sufficient to answer the question. For example, when n = 3, the remainder when n is divided by 2 is 1. However, when n = 16, the remainder when n is divided by 2 is 0.

Statement Two Alone:

n + 2 is a multiple of 7.

Statement two alone is not sufficient to answer the question. For instance, when n = 5, the remainder when n is divided by 2 is 1. However, when n = 12, the remainder when n is divided by 2 is 0.

Statements One and Two Together:

Let’s list out possible values of n from statement one:

From statement one:

n = 3, 16, 29, 42, 55, 68

We see that from the above list, 68 (since 68 + 2 = 70 is a multiple of 7) fulfills both statements one and two and provides a remainder of 0 when divided by 2.

To determine the next value in the list, we can use the least common multiple of 13 and 7, which is 13 x 7 = 91. Thus, the next number that could be n is 68 + (7 x 13) = 159. Since 159 has a remainder of 1 when divided by 2, the two statements together are still not sufficient to answer the question.

Answer: E
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