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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.

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.

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.

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

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18 Jun 2010, 06:46

Expert's post

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)).

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 - \(0\leq{r}<d\). _________________

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

Answer C.

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 ?

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

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.

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

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06 Mar 2011, 16:41

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.

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