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When the positive integer k is divided by the positive integer n , the remainder is 11. If k/n = 81.2 , what is the value of n?

A. 9 B. 20 C. 55 D. 70 E. 81

Can someone break this down for me please?

Here is a response from Beat The GMAT (/tricky-remainder-problem-t270176.html)

Before I past the question, I am confused with how the user came up with K = 81n +11. Where did the .2 go for this representation?

"If, when k is divided by n, the remainder is 11, we could say that some multiple of n plus 11 equals k: xn + 11 = k

If k/n = 81.2, that means that "some multiple of n" (aka the quotient) is 81, and the remainder is represented by the 0.2.

k = 81.2n

and

k = 81n + 11

Now, we can simply set these expressions equal to each other, since they're both equal to k:

81.2n = 81n + 11 0.2n = 11 n = 55"

Thank you for your help!!

If \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(y =divisor*quotient+remainder= xq + r\) and \(0\leq{r}<x\).

For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since \(15 = 6*2 + 3\).

Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.

We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.

Re: When the positive integer k is divided by the positive integ [#permalink]

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13 Aug 2014, 04:59

Bunuel wrote:

If \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(y =divisor*quotient+remainder= xq + r\) and \(0\leq{r}<x\).

For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since \(15 = 6*2 + 3\).

Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.

We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.

Answer: C.

How did you know to create K = 81n + 11? How did you know to leave out the 0.2? I am confused as heck by this!

If \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(y =divisor*quotient+remainder= xq + r\) and \(0\leq{r}<x\).

For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since \(15 = 6*2 + 3\).

Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.

We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.

Answer: C.

How did you know to create K = 81n + 11? How did you know to leave out the 0.2? I am confused as heck by this!

Please follow the links given in my post. You need to brush up fundamentals.
_________________

Here's an approach that's based on Number Properties and a bit of "brute force" math:

We're told that K and N are both INTEGERS.

Since K/N = 81.2, we can say that K = 81.2(N)

N has to "multiply out" the .2 so that K becomes an INTEGER. With the answers that we have to work with, N has to be a multiple of 5. Eliminate A and E.

With the remaining answers, we can TEST THE ANSWERS and find the one that fits the rest of the info (K/N = 81.2 and K/N has a remainder of 11)

Answer B: If N = 20, then K = 1624; 1624/20 has a remainder of 4 NOT A MATCH Answer C: If N = 55, then K = 4466; 4466/55 has a remainder of 11 MATCH.

Re: When the positive integer k is divided by the positive integ [#permalink]

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29 Dec 2014, 17:18

1

This post received KUDOS

Yes, I am aware that my answer goes after the 800-scorer Rich, but I´ll try my best!

Once you really understood the pattern in which Gmat works with this type of problem, it all boils down to a very simple system of 3 equations and 3 variables. That´s the "trick" you should know.

Here is my reasoning:

The System of Equations 1st Equation: k = nq + 11 --> k/n = q + 11/n 2nd Equation: k/n = 81.2 --> k/n = 81 + 1/5 3rd Equation: q = 81

Plug in q (from equation 3) in equation 1 k/n = 81 + 11/n

Re: When the positive integer k is divided by the positive integ [#permalink]

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31 May 2016, 05:10

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Re: When the positive integer k is divided by the positive integ [#permalink]

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20 Jul 2017, 07:17

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: When the positive integer k is divided by the positive integ [#permalink]

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28 Sep 2017, 04:00

Bunuel wrote:

joaomario wrote:

When the positive integer k is divided by the positive integer n , the remainder is 11. If k/n = 81.2 , what is the value of n?

A. 9 B. 20 C. 55 D. 70 E. 81

Can someone break this down for me please?

Here is a response from Beat The GMAT (/tricky-remainder-problem-t270176.html)

Before I past the question, I am confused with how the user came up with K = 81n +11. Where did the .2 go for this representation?

"If, when k is divided by n, the remainder is 11, we could say that some multiple of n plus 11 equals k: xn + 11 = k

If k/n = 81.2, that means that "some multiple of n" (aka the quotient) is 81, and the remainder is represented by the 0.2.

k = 81.2n

and

k = 81n + 11

Now, we can simply set these expressions equal to each other, since they're both equal to k:

81.2n = 81n + 11 0.2n = 11 n = 55"

Thank you for your help!!

If \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(y =divisor*quotient+remainder= xq + r\) and \(0\leq{r}<x\).

For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since \(15 = 6*2 + 3\).

Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.

We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.

When the positive integer k is divided by the positive integer n , the remainder is 11. If k/n = 81.2 , what is the value of n?

A. 9 B. 20 C. 55 D. 70 E. 81

Can someone break this down for me please?

Here is a response from Beat The GMAT (/tricky-remainder-problem-t270176.html)

Before I past the question, I am confused with how the user came up with K = 81n +11. Where did the .2 go for this representation?

"If, when k is divided by n, the remainder is 11, we could say that some multiple of n plus 11 equals k: xn + 11 = k

If k/n = 81.2, that means that "some multiple of n" (aka the quotient) is 81, and the remainder is represented by the 0.2.

k = 81.2n

and

k = 81n + 11

Now, we can simply set these expressions equal to each other, since they're both equal to k:

81.2n = 81n + 11 0.2n = 11 n = 55"

Thank you for your help!!

If \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(y =divisor*quotient+remainder= xq + r\) and \(0\leq{r}<x\).

For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since \(15 = 6*2 + 3\).

Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.

We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.

"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."

I was wondering, why can't q = 0.2 and 81=11/n ?

The point is that the quotient must be an integer.

For example, 15 divided by 6 gives the quotient of 2 and the remainder of 3: 15 = 2*6 + 3. The quotient (2 in our case) is the greatest whole number of times a divisor (6 in our case) may be subtracted from a dividend (15) without the remainder becoming negative.
_________________

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