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# When 10 is divided by the positive integer n, the remainder

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

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15 Sep 2008, 14:33
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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
[Reveal] Spoiler: OA
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Re: Number properties question from QR 2nd edition PS 164 [#permalink]

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18 Jun 2010, 01:58
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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.

Hope it's clear.
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16 Sep 2008, 07:18
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10=NK+N-4; assume K=1

10=2N-4

14=2N. N has to be a multiple of 7...

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

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16 Jul 2010, 07:14
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nonameee wrote:
Quote:
remainder is always non-negative

Bunuel, I have to disagree with you on that:

http://en.wikipedia.org/wiki/Remainder

This has nothing to do with GMAT.

GMAT Prep definition of the remainder:

If $$a$$ and $$d$$ are positive integers, there exists unique integers $$q$$ and $$r$$, such that $$a = qd + r$$ and $$0\leq{r}<d$$. $$q$$ is called a quotient and $$r$$ is called a remainder.

Also EVERY GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

So trust me: remainder is always non-negative and less than divisor for GMAT - $$0\leq{r}<d$$.
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Re: Number properties question from QR 2nd edition PS 164 [#permalink]

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06 Mar 2011, 16:54
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Spidy001 wrote:
Bunuel,

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

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09 Jan 2013, 04:24
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fozzzy wrote:
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.

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.

Hope it's clear.
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15 Sep 2008, 17:08
back solving is easier for this one.

10/7 gives a remainder of 3. n-4 = 7-4 = 3
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17 Sep 2008, 07:04
vksunder wrote:
fresinha12 - I did the same way as you had described. But is it safe to assume that K=1?

well, since we can never have the denominator to be zero, otherwise the fraction will be undefined. so it makes sense to start off with k=1. If that doesn't work, then you just have to keep increasing the value of k until you can match your answer with the correct answer choice.
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18 Sep 2008, 00:18
I approached as follows.

10 = nx + n-4 for x = 0,1,2,3,4......
or, n(x+1) = 14
or, n = 14/(x+1)

For x = 0, n = 14, for x = 1, n = 7, x cannot be 2,3,4,5.
For x = 6, n = 2. x cannot be greater than 6.

Hence, possible values of n are 14, 7, 2. Answer choice has 7. Hence, 7 is the answer.
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Number properties question from QR 2nd edition PS 164 [#permalink]

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18 Jun 2010, 00:12
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?
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Re: Number properties question from QR 2nd edition PS 164 [#permalink]

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18 Jun 2010, 00:56
As per my approach, it is easy to reach the solution by going thorough each one of the options.
You can eliminate 12,8,4 and 3 at one look. Then you just need to check for 7. It took me less than 1 minute to get to the answer. So that should be fine I guess.
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Re: Number properties question from QR 2nd edition PS 164 [#permalink]

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18 Jun 2010, 06:29
It says that the remainder when you divide 10 by n is n-4

This basically can be translated into the following statement algebraically:

$$10 = kn + (n-4)$$

This is simplified as follows:

$$10 = kn + n -4 = n *(k+1) - 4$$

Further simplifying:

$$10 + 4 = n*(k+1) 14 = n*(k+1) 7*2 = n*(k+1)$$

So n can be 7 or 2.

Only 7 is listed as an option here, so the answer is C. Hope this helps!

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

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18 Jun 2010, 06:46
whiplash2411 wrote:
It says that the remainder when you divide 10 by n is n-4

This basically can be translated into the following statement algebraically:

$$10 = kn + (n-4)$$

This is simplified as follows:

$$10 = kn + n -4 = n *(k+1) - 4$$

Further simplifying:

$$10 + 4 = n*(k+1) 14 = n*(k+1) 7*2 = n*(k+1)$$

So n can be 7 or 2.

Only 7 is listed as an option here, so the answer is C. Hope this helps!

$$n$$ cannot be 2 as in this case $$remainder =n-4=-2<0$$ and remainder is always non-negative (also notice that 10/2 has no remainder and n-4=-2, though n can also be 14 --> 10=14*0+(14-4)).

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

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18 Jun 2010, 06:52
Oh, yeah, that's right. I just saw the 7 and 2, and looked at the answer choices and chose 7.

Thanks, Bunuel. Your explanation will come in handy in case both 2 and 7 were listed as answer choices!
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Re: Number properties question from QR 2nd edition PS 164 [#permalink]

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16 Jul 2010, 04:22
Quote:
remainder is always non-negative

Bunuel, I have to disagree with you on that:

http://en.wikipedia.org/wiki/Remainder
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Re: Number properties question from QR 2nd edition PS 164 [#permalink]

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16 Jul 2010, 14:00
Thanks for clarification. But you can use that property (negative remainder) to solve remainder problems (as it has been done in several posts).
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05 Mar 2011, 01:24
If division by n leaves reminder. Then
i.e. Dividend - Remainder is a multiple of divider.
Here 10 -(n-4) must be a multiple of n.

Or Is [10 - (n-4)] / n = integer?

Now plug in the values of n from the options.

A - n-4 will give negative remainder. Illogical
B - (10-0)/4 is not integer
C - (10-3)/7 is integer
D - (10-4)/8 is not integer
E - (10-8)/12 is not integer

Baten80 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
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05 Mar 2011, 01:44
$$10=nQ+n-4$$ where Q is the quotient

$$n(Q+1)=14$$, where n and Q are both integers.

Factors of 14;
n*(Q+1)
1*14; n=1, Q=13; Not possible because 1 won't leave any remainder with 10
2*7; n=2, Q=6; Not possible because 2 won't leave any remainder with 10
7*2; n=7, Q=1; Possible
14*1; n=14, Q=0; Possible

So; n can be 7 or 14.

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

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06 Mar 2011, 15:04
Nice explanation there Bunuel.

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).

$$10=nq+n-4$$ --> $$14=n(q+1)$$ --> $$n$$ is an factor of 14 and $$\geq{4}$$ --> $$n$$ can be 7 or 14.

Hope it's clear.
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Re: Number properties question from QR 2nd edition PS 164 [#permalink]

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06 Mar 2011, 15:48
I thought of the same , why cant a remainder be negative?

I guess in some cases , as Bunel is suggesting we need to make an assumption that we are dealing with just positive integers.

nonameee wrote:
Quote:
remainder is always non-negative

Bunuel, I have to disagree with you on that:

http://en.wikipedia.org/wiki/Remainder

Last edited by Spidy001 on 06 Mar 2011, 16:27, edited 1 time in total.
Re: Number properties question from QR 2nd edition PS 164   [#permalink] 06 Mar 2011, 15:48

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