When the positive integer k is divided by the positive integ

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.

Similar questions to practice:
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if-s-and-t-are-positive-integers-such-that-s-t-64-12-which-135190.html

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rule 3. Thank you.

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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

Rewrite Equations 1 and 2
k/n – 81 = 11/n
k/n – 81 = 1/5

Set Equalities in Equations 1 and 2
11/n = 1/5
n = 55

There are many ways to find the system of equations. I just found this approach "elegant".

Hope it helps!

If K/N is 81,2 and the remainder is 11 we can say that 11=0,2n
11/0,2=55
N=55

Answer C

Hi joaomario,

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.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hi Bunuel,

"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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We can create two remainder equations:

k/n = 81 + 2/10

k/n = 81 + 1/5

and

k/n = 81 + 11/n

Thus:

1/5 = 11/n

n = 55

Answer: C
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could someone please explain the logic behind writing 81.2 in form 81 + 0.2 or 81 + 1/5?

Hi davidbeckham,

GMAT questions can almost always be solved in more than one way (including how you might choose to work through the 'math steps' involved). Sometimes 'converting' information from one format to another can make it easier for you to answer the given question (depending on how you like to "see" your math work).

Some of the explanations in this thread 'convert' 81.2 in the way that you describe, but that is NOT a necessary step to answering this question. Working in that way is just one option for dealing with the fact that both K and N must be INTEGERS and K/N has a remainder of 11.

GMAT assassins aren't born, they're made,
Rich

Everything you learn about remainders will be about integers, so you generally would always want to rewrite a question like this using integers instead of decimals. From the definition of a quotient and a remainder, then when we divide integer n by integer d,

n = qd + r

where integer q is the quotient, and integer r is the remainder. The remainder must be between 0 and d-1 inclusive, so r < d is always true. If you divide by d on both sides, you get

n/d = q + (r/d)

Since r < d, then r/d must be between 0 and 1. So when you divide n by d, the quotient q is the whole number part of the result, and r/d is the decimal or fractional part of the result. If we use that for this question (and I'm going to change the letters, because whoever designed this question did something profoundly annoying -- they're using "n" where "d" is normally used in a quotient/remainder definition, and using "k" where "n" would normally be used) :

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

then 81 is the whole number part of the result, so is the quotient, and 0.2 = 1/5 is the decimal part of the result, so equals r/d. Since r = 11, and r/d = 1/5, we have 11/d = 1/5 and d = 55.

So if you replace 0.2 with a fraction of integers, and you know how to use the definition of a remainder, it's a one-line solution. It's true that you can solve this question in other ways, as the post above mine suggests, but if you're doing something like testing answers here, you're spending far more time on this question than you need to.
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this question is not difficult but ya , a bit tricky to understand when you see this question at first.
so ,
K/N = 11 i.e. K= NQ+11
K/N =81.2= K=N*81+0.2 for example : 16/3= 3*5+1
now ,
K= 81N+11
81.2N=81N+11
N=55.

Partly a bit of fun. Hope it helps someone out there! This tutor is going old-school for this! (Gosh, his writing is terrible すみません!) *technically this topic was done yesterday - It's now past midnight where I am!

https://ibb.co/2K4Pckp
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We're told \(\frac{k}{n} = 81.2\)

We can rewrite 81.2 as \(81 \frac{1}{5}\)

That 1/5 represents a remainder of 11.

Therefore, 5/5 = 55.

Answer is C.
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Video solution from Quant Reasoning starts at 5:45:
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basshead How did you get 5/5 is 55? I got the 1/5 is 11, but how the 5/5? I'm a bit lost & slow lol

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