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The remainder when N is divided by 18 is 16. Given that N is [#permalink]
26 Apr 2013, 02:28

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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 N/4 is divided by 18?

Re: The remainder when N is divided by 18 is 16 [#permalink]
26 Apr 2013, 03:18

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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!) _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: The remainder when N is divided by 18 is 16 [#permalink]
26 Apr 2013, 03:46

Zarrolou wrote:

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!)

Thank you so much for solution and kudos!

It took me some time to find the nice solution. I will post how I see the solution later here. I'm just waiting for possible other comments. _________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

Re: The remainder when N is divided by 18 is 16 [#permalink]
26 Apr 2013, 04:49

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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 .

Re: The remainder when N is divided by 18 is 16. Given that N is [#permalink]
27 Apr 2013, 01:10

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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. _________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

Re: The remainder when N is divided by 18 is 16. Given that N is [#permalink]
29 Jul 2014, 00:30

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Re: The remainder when N is divided by 18 is 16. Given that N is [#permalink]
11 Oct 2015, 05:01

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