The remainder when N is divided by 18 is 16. Given that N is

So, my solution. Just a little bit different from the previous.


The remainder when \(N\) is divided by 18 is 16 means that \(N=18q+16\) for some integer \(q\).
\(N\) is a multiple of 28 means that \(N=28s\) for some integer \(s\).

We need to find the remainder when \(\frac{N}{4}\) is divided by 18.

On one hand \(\frac{N}{4}=7s\), on the other hand \(\frac{N}{4}=\frac{9q}{2}+4\). Since \(7s=\frac{9q}{2}+4\) and \(s\) is an integer, \(q\) must be even.

So, \(\frac{N}{4}=9k+4\) for some integer \(k\).
If\(k\) is even (\(k=2n\) for some integer \(n\)) the remainder when \(\frac{N}{4}\) is divided by 18 is 4 (\(\frac{N}{4}=9*2n+4=18n+4\)).
If \(k\) is odd (\(k=2n+1\) for some integer \(n\)) the remainder when \(\frac{N}{4}\) is divided by 18 is 13 (\(\frac{N}{4}=9(2n+1)+4=18n+13\)).

So, there two possible values for the remainder 4 and 13.
The answer is B.

The remainder when N is divided by 18 is 16. Given that N is a multiple of 28, how many integers between 0 and 18 inclusive could be the remainder when
\frac{N}{4} is divided by 18?

Let N = 28x
so 28x = 18y + 16 or 18z - 2 both are equivalent .
so 28x = 18z -2 according to statement mentioned .

Now remainder when N/4 is divided by 18
let remainder be R
Let N/4 = 18q + R
Substituting N = 28x = 18z-2 we get
18z -2 = 72q + 4R
therefore R = (18(z - 4q)-2)/4 = (9(z - 4q ) - 2 ) /2 = (9*someinteger - 1) /2
If a number is divided by 18 so remainder is between 1 and 17 .
Substituting integer values we get :
(9*1 -1)/2 = 4 possible remainder
(9*2 -1 )/2 = 8.5 not possible
(9*3 -1 )/2 = 13 possible
(9*4 -1 )/2 = 17.5 not possible

Thus we get only 2 possible values for remainder i.e 4 and 13 hence answer is 2 .

The remainder when N is divided by 18 is 16, translated : \(N=18k+16\)
\(\frac{N}{4}\) is divided by 18 means what is the remainder of \(\frac{N}{4*18}\)?
Given that N is a multiple of 28, translated: \(N=28m\)

\(\frac{N}{4*18}\) with \(N=28m\) is \(\frac{28m}{4*18}\) or \(\frac{7m}{18}\) and its "form" can be written as \(7m=18q+R\) ( or 14m=36q+2R, this will be useful later)

Going back to the first equation \(N=18k+16\) = \(28m=18k+16\) = \(14m=9k+8\). From the equation before is its "useful" form 14m=36q+2R
so puttin them together \(9k+8=36q+2R\) all the numbers k,q,R must be integer

\(8-2R=36q-9k\)
if q and r are 0 \(8-2R=0\) so \(R=4\) value #1
the other possible value of R (because must be positive, it's a reminder) will be in the case 9k>36q
The difference \(36q-9k\) can be (36-45) = -9 but \(8-2R=-9\) means R=17/2 no integer
difference -18 => R = 5 value #2
difference -27 => R = 33/2 no integer
difference -36 => R=21 out of range 0,18
We can stop here bigger differences mean R out of 0,18 range

2 values, B
(I am not sure of my method though, the Master Mind could help here and +1 to the question! it took me 10 minutes to came up with a solution!)

Answer is B

N= 18i + 16, and N is multiple of 28
N can have values, = 196,376,556,736 .. and so on
N/4 is 49,94,139,184.....
remainder when divided by 18 gives ... 13,4,13,4 resp.

Therefore only 2 values possible

Answer is B
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Is this a GMAT question?

Hi smyarga

I did not understand this step and thus nothing after it:

So, \(\frac{N}{4}=9k+4\) for some integer \(k\).

How did the 9q/2 changed into 9k?

Tagging others just in case
Bunuel chetan2u GMATinsight ScottTargetTestPrep IanStewart yashikaaggarwal

Thank you

Posted from my mobile device

That solution was just noting that q/2 must be an integer, so to make things easier to look at, it replaced "q/2" with the integer "k".

Someone asked above if this was a GMAT question, and the answer is "no."

Hi

As Ian too has pointed out, 9q/2 is an integer.
N being a multiple of 28 is just to tell you that N is a multiple of 4.
So 18q+16 has to be divisible by 4. This means q has to be even say 2k
18q+16=18*2k+16, which when divided by 4 gives 9k+4.

Now 9k+4 has to be divided by 18.
When k is odd, 9k will leave 9 as remainder => R =9+4=13.
When k is even, 9k will leave 0 as remainder => R =0+4=4

So two possible values
_________________

We know N/4 = 9q/2 + 4 and we know q is even. Since q is even, we can express q as q = 2k for some integer k. Substituting q by 2k, we obtain:


N/4 = 9(2k)/2 + 4 = 9k + 4

That's how the expression 9k + 4 is obtained. Now, the integer k is either even or odd. If k is even, k can be expressed as k = 2n for some integer n and if k is odd, k can be expressed as k = 2m + 1 for some integer m. In one case, we obtain the equality N = 18n + 4 and in the other case, we obtain N = 18m + 13.
At this point, we need to recall the definition of division with remainders: when a is divided by b, we say that the quotient is q and the remainder is r if a = bq + r and 0 ≤ r < b.

Using the equality N = 18n + 4, we see that r = 4 satisfies 0 ≤ 4 < 18; so that's the remainder when N is divided by 18. Similarly, the equality N = 18m + 13 tells us that the remainder when N is divided by 18 is 13. Thus, there are two possible values for the remainder when N is divided by 18.
_________________

smyarga
Given: The remainder when N is divided by 18 is 16.
Asked: Given that N is a multiple of 28, how many integers between 0 and 18 inclusive could be the remainder when N/4 is divided by 18?
N = 18k + 16 = 28m
9k + 8 = 14m
k = (14m-8)/9 = 2(7m-4)/9

m = {7, 16, 25, 34, 43, 52, 61, 70,...}
m = 7 + 9s

Case 1: s = 2p
m = 7 + 18p
N/4 = 7m = 7(7+18p) = 49 + 7*18p
The remainder when N/4 is divided by 18 = 13

Case 2: s = 2p + 1
m = 7 + 18p + 9
m = 18p + 16
N/4 = 7*18p + 112
The remainder when N/4 is divided by 18 = 4

The remainder when N/4 is divided by 18 = {13,4}

IMO B
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