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I know in this case we don't have to make any assumption, because the question clearly states these are two positive integers.

i was referring more to scenarios like negative number division

-25 /7

-25 = 7(-3)+(-4)

Here remainder is -4 which is negative.

so lets say if question is like x,y are integers x/y . we cannot generalize and say remainder >=0 ,unless we assume that we are only talking about positive integers.

Two things:

1. Every GMAT divisibility question will tell you in advance that any unknowns represent positive integers. 2. A remainder is a non-negative integer by definition (at least on the GMAT).

Anyway you are still wrong when calculating -25/7, it should be: -25=(-4)*7+3, so remainder=3>0.

TO SUMMARIZE, DON'T WORRY ABOUT NEGATIVE DIVIDENDS, DIVISORS OR REMAINDERS ON THE GMAT.
_________________

Re: Number properties question from QR 2nd edition PS 164 [#permalink]

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09 Jan 2013, 03:02

Bunuel wrote:

jpr200012 wrote:

When 10 is divided by the positive integer n, the remainder is n-4. Which of the following could be the value of n?

A. 3 B. 4 C. 7 D. 8 E. 12

My strategy was to create lists below: n = 3, 4, 7, 8, 12 n-4 = -1(becomes 9), 0, 3, 4, 8 n/10 = R? = 3, 4, 7, 8, 4

There is no match between n-4 and n/10's R.

The solution uses 14 = ..., but I don't understand how they are using 14. Should the question have said a multiple of one of these numbers?

Algebraic approach:

THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

Original question says that when 10 is divided by the positive integer n, the remainder is n-4, so \(10=nq+(n-4)\) and also \(n-4\geq{0}\) or \(n\geq{4}\) (remainder must be non-negative).

\(10=nq+n-4\) --> \(14=n(q+1)\) --> as \(14=1*14=2*7\) and \(\geq{4}\) then --> \(n\) can be 7 or 14.

Answer: C.

Hope it's clear.

So in this step are we substituting q=0,1 etc or is it something else?
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When 10 is divided by the positive integer n, the remainder is n-4. Which of the following could be the value of n?

A. 3 B. 4 C. 7 D. 8 E. 12

My strategy was to create lists below: n = 3, 4, 7, 8, 12 n-4 = -1(becomes 9), 0, 3, 4, 8 n/10 = R? = 3, 4, 7, 8, 4

There is no match between n-4 and n/10's R.

The solution uses 14 = ..., but I don't understand how they are using 14. Should the question have said a multiple of one of these numbers?

Algebraic approach:

THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

Original question says that when 10 is divided by the positive integer n, the remainder is n-4, so \(10=nq+(n-4)\) and also \(n-4\geq{0}\) or \(n\geq{4}\) (remainder must be non-negative).

\(10=nq+n-4\) --> \(14=n(q+1)\) --> as \(14=1*14=2*7\) and \(\geq{4}\) then --> \(n\) can be 7 or 14.

Answer: C.

Hope it's clear.

So in this step are we substituting q=0,1 etc or is it something else?

Not entirely so.

From \(10=nq+n-4\):

Re-arrange: \(14=nq+n\); Factor out n: \(14=n(q+1)\).

So we have that the product of two positive integers (n and q+1) equals 14. 14 can be written as the product of two positive integers only in 2 way: 14=1*14 and 14=2*7. Now, since \(n\geq{4}\) then \(n\) can be 7 or 14.

Re: Number properties question from QR 2nd edition PS 164 [#permalink]

Show Tags

21 Sep 2013, 08:10

Bunuel wrote:

jpr200012 wrote:

When 10 is divided by the positive integer n, the remainder is n-4. Which of the following could be the value of n?

A. 3 B. 4 C. 7 D. 8 E. 12

My strategy was to create lists below: n = 3, 4, 7, 8, 12 n-4 = -1(becomes 9), 0, 3, 4, 8 n/10 = R? = 3, 4, 7, 8, 4

There is no match between n-4 and n/10's R.

The solution uses 14 = ..., but I don't understand how they are using 14. Should the question have said a multiple of one of these numbers?

Algebraic approach:

THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

Original question says that when 10 is divided by the positive integer n, the remainder is n-4, so \(10=nq+(n-4)\) and also \(n-4\geq{0}\) or \(n\geq{4}\) (remainder must be non-negative).

\(10=nq+n-4\) --> \(14=n(q+1)\) --> as \(14=1*14=2*7\) and \(\geq{4}\) then --> \(n\) can be 7 or 14.

Answer: C.

Hope it's clear.

Hi Bunuel, I also considered n=7,14 as the only two options since the remainder has to be non-negative. However, the following official explanation (Quant Review 2nd edition, PS 164) confused me:

"10 = qn + (n- 4}. So, 14 = qn + n = n(q + 1). This means that n must be a factor of 14 and so n= 1, n = 2, n = 7, or n = 14 since n is a positive integer and the only positive integer factors of 14 are 1, 2, 7, and 14. The only positive integer factor of 14 given in the answer choices is 7."

Here n=1 and n=2 are considered as possible values for n even though that will make the remainder n-4 negative.

When 10 is divided by the positive integer n, the remainder is n-4. Which of the following could be the value of n?

A. 3 B. 4 C. 7 D. 8 E. 12

My strategy was to create lists below: n = 3, 4, 7, 8, 12 n-4 = -1(becomes 9), 0, 3, 4, 8 n/10 = R? = 3, 4, 7, 8, 4

There is no match between n-4 and n/10's R.

The solution uses 14 = ..., but I don't understand how they are using 14. Should the question have said a multiple of one of these numbers?

Algebraic approach:

THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

Original question says that when 10 is divided by the positive integer n, the remainder is n-4, so \(10=nq+(n-4)\) and also \(n-4\geq{0}\) or \(n\geq{4}\) (remainder must be non-negative).

\(10=nq+n-4\) --> \(14=n(q+1)\) --> as \(14=1*14=2*7\) and \(\geq{4}\) then --> \(n\) can be 7 or 14.

Answer: C.

Hope it's clear.

Hi Bunuel, I also considered n=7,14 as the only two options since the remainder has to be non-negative. However, the following official explanation (Quant Review 2nd edition, PS 164) confused me:

"10 = qn + (n- 4}. So, 14 = qn + n = n(q + 1). This means that n must be a factor of 14 and so n= 1, n = 2, n = 7, or n = 14 since n is a positive integer and the only positive integer factors of 14 are 1, 2, 7, and 14. The only positive integer factor of 14 given in the answer choices is 7."

Here n=1 and n=2 are considered as possible values for n even though that will make the remainder n-4 negative.

These values are considered solely based on 14=n(q+1).
_________________

Re: Number properties question from QR 2nd edition PS 164 [#permalink]

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21 Sep 2013, 08:47

Bunuel wrote:

These values are considered solely based on 14=n(q+1).

So, I guess the official explanation is incomplete in the sense that it doesn't take into account the properties of remainders. I am surprised n-4>=0 wasn't taken into account but hope it is a mistake rather than the possibility that the remainder rule is not strictly applicable.

Re: Number properties question from QR 2nd edition PS 164 [#permalink]

Show Tags

25 Sep 2013, 07:09

Bunuel wrote:

jpr200012 wrote:

When 10 is divided by the positive integer n, the remainder is n-4. Which of the following could be the value of n?

A. 3 B. 4 C. 7 D. 8 E. 12

My strategy was to create lists below: n = 3, 4, 7, 8, 12 n-4 = -1(becomes 9), 0, 3, 4, 8 n/10 = R? = 3, 4, 7, 8, 4

There is no match between n-4 and n/10's R.

The solution uses 14 = ..., but I don't understand how they are using 14. Should the question have said a multiple of one of these numbers?

Algebraic approach:

THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

Original question says that when 10 is divided by the positive integer n, the remainder is n-4, so \(10=nq+(n-4)\) and also \(n-4\geq{0}\) or \(n\geq{4}\) (remainder must be non-negative).

\(10=nq+n-4\) --> \(14=n(q+1)\) --> as \(14=1*14=2*7\) and \(\geq{4}\) then --> \(n\) can be 7 or 14.

Answer: C.

Hope it's clear.

I got stuck when I got to 14=n(q+1) - so do we just completely ignore the 'q'? and why do we ignore the 'q'? can't q be something like 13 and 'n' becomes any random number? what am I missing here?

When 10 is divided by the positive integer n, the remainder is n-4. Which of the following could be the value of n?

A. 3 B. 4 C. 7 D. 8 E. 12

My strategy was to create lists below: n = 3, 4, 7, 8, 12 n-4 = -1(becomes 9), 0, 3, 4, 8 n/10 = R? = 3, 4, 7, 8, 4

There is no match between n-4 and n/10's R.

The solution uses 14 = ..., but I don't understand how they are using 14. Should the question have said a multiple of one of these numbers?

Algebraic approach:

THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

Original question says that when 10 is divided by the positive integer n, the remainder is n-4, so \(10=nq+(n-4)\) and also \(n-4\geq{0}\) or \(n\geq{4}\) (remainder must be non-negative).

\(10=nq+n-4\) --> \(14=n(q+1)\) --> as \(14=1*14=2*7\) and \(\geq{4}\) then --> \(n\) can be 7 or 14.

Answer: C.

Hope it's clear.

I got stuck when I got to 14=n(q+1) - so do we just completely ignore the 'q'? and why do we ignore the 'q'? can't q be something like 13 and 'n' becomes any random number? what am I missing here?

We don't ignore q, we are just not interested in it. q is a quotient, so is a non-negative integer, thus we have 14=n(q+1)=integer*integer --> both multiples are factors of 14.

Re: Number properties question from QR 2nd edition PS 164 [#permalink]

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26 Sep 2013, 01:56

Bunuel wrote:

bulletpoint wrote:

I got stuck when I got to 14=n(q+1) - so do we just completely ignore the 'q'? and why do we ignore the 'q'? can't q be something like 13 and 'n' becomes any random number? what am I missing here?

We don't ignore q, we are just not interested in it. q is a quotient, so is a non-negative integer, thus we have 14=n(q+1)=integer*integer --> both multiples are factors of 14.

Does this make sense?

why do both 'n' and '(q+1)' have to be factors of 14? if 'q+1' is a factor of 14, then 'n' need not be a factor of 14 for the equation 14=n(q+1) to be true, right?

or is it that for questions of these types - since we are only interested in what 'n' is - we just completely ignore the '(q+1)' part?

EDIT: Just took a look at what you said again and I think I get it. Please correct me if I'm wrong:

14=n(q+1) means 'n' OR '(q+1)' can equal 1,2,7,14 to make the equation true, and since 'n' has to be greater or equal to 4 because remainder must be non-negative, it can only be true that 'n' equals 7 or 14, and because the answer only has 7, this would be the correct answer.

Last edited by bulletpoint on 26 Sep 2013, 02:00, edited 1 time in total.

I got stuck when I got to 14=n(q+1) - so do we just completely ignore the 'q'? and why do we ignore the 'q'? can't q be something like 13 and 'n' becomes any random number? what am I missing here?

We don't ignore q, we are just not interested in it. q is a quotient, so is a non-negative integer, thus we have 14=n(q+1)=integer*integer --> both multiples are factors of 14.

Does this make sense?

why do both 'n' and '(q+1)' have to be factors of 14? if 'q+1' is a factor of 14, then 'n' need not be a factor of 14 for the equation 14=n(q+1) to be true, right?

or is it that for questions of these types - since we are only interested in what 'n' is - we just completely ignore the '(q+1)' part?

Again we do NOT ignore q+1.

Next, 14 = n(q+1) = integer*integer: 14/n = q+1 = integer --> n is a factor of 14. 14/(q+1) = n = integer --> q+1 is a factor of 14.
_________________

Re: Number properties question from QR 2nd edition PS 164 [#permalink]

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23 Oct 2013, 17:42

Bunuel wrote:

jpr200012 wrote:

When 10 is divided by the positive integer n, the remainder is n-4. Which of the following could be the value of n?

A. 3 B. 4 C. 7 D. 8 E. 12

My strategy was to create lists below: n = 3, 4, 7, 8, 12 n-4 = -1(becomes 9), 0, 3, 4, 8 n/10 = R? = 3, 4, 7, 8, 4

There is no match between n-4 and n/10's R.

The solution uses 14 = ..., but I don't understand how they are using 14. Should the question have said a multiple of one of these numbers?

Algebraic approach:

THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

Original question says that when 10 is divided by the positive integer n, the remainder is n-4, so \(10=nq+(n-4)\) and also \(n-4\geq{0}\) or \(n\geq{4}\) (remainder must be non-negative).

\(10=nq+n-4\) --> \(14=n(q+1)\) --> as \(14=1*14=2*7\) and \(\geq{4}\) then --> \(n\) can be 7 or 14.

Answer: C.

Hope it's clear.

could you clarify the highlighted portion? is n being 7 because 14=2*7?

When 10 is divided by the positive integer n, the remainder is n-4. Which of the following could be the value of n?

A. 3 B. 4 C. 7 D. 8 E. 12

My strategy was to create lists below: n = 3, 4, 7, 8, 12 n-4 = -1(becomes 9), 0, 3, 4, 8 n/10 = R? = 3, 4, 7, 8, 4

There is no match between n-4 and n/10's R.

The solution uses 14 = ..., but I don't understand how they are using 14. Should the question have said a multiple of one of these numbers?

Algebraic approach:

THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

Original question says that when 10 is divided by the positive integer n, the remainder is n-4, so \(10=nq+(n-4)\) and also \(n-4\geq{0}\) or \(n\geq{4}\) (remainder must be non-negative).

\(10=nq+n-4\) --> \(14=n(q+1)\) --> as \(14=1*14=2*7\) and \(\geq{4}\) then --> \(n\) can be 7 or 14.

Answer: C.

Hope it's clear.

could you clarify the highlighted portion? is n being 7 because 14=2*7?

Yes, we know that \(n\geq{4}\) and \(14=n*(positive \ integer)\). Now, \(14=1*14=2*7\), thus \(n\) can be 7 or 14.

Re: When 10 is divided by the positive integer n, the remainder [#permalink]

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17 May 2016, 13:02

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Re: When 10 is divided by the positive integer n, the remainder [#permalink]

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04 Dec 2016, 01:32

In this Question,Back-solving is Straight forward. 7 is a clear Hit. Also observe there are two values of n that will satisfy the given conditions=> {7,14}

Hence C _________________

Give me a hell yeah ...!!!!!

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Re: When 10 is divided by the positive integer n, the remainder
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04 Dec 2016, 01:32

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