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The remainder when N is divided by 18 is 16. Given that N is [#permalink]
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26 Apr 2013, 03:28
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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? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 Just nice problem from http://www.mualphatheta.org/National_Co ... Tests.aspxI know some ways how to solve it quickly. May be someone knows nicer way of solution. Thanks:)
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Re: The remainder when N is divided by 18 is 16. Given that N is [#permalink]
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27 Apr 2013, 02:10
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.
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Re: The remainder when N is divided by 18 is 16 [#permalink]
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26 Apr 2013, 04:18
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 \(82R=36q9k\) if q and r are 0 \(82R=0\) so \(R=4\) value #1the other possible value of R (because must be positive, it's a reminder) will be in the case 9k>36q The difference \(36q9k\) can be (3645) = 9 but \(82R=9\) means R=17/2 no integer difference 18 => R = 5 value #2difference 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!)
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Re: The remainder when N is divided by 18 is 16 [#permalink]
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26 Apr 2013, 04:46
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
\(82R=36q9k\) if q and r are 0 \(82R=0\) 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 \(36q9k\) can be (3645) = 9 but \(82R=9\) 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.
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Re: The remainder when N is divided by 18 is 16 [#permalink]
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26 Apr 2013, 05:49
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 = 18z2 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 .



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Re: The remainder when N is divided by 18 is 16. Given that N is [#permalink]
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18 Aug 2017, 03:36
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 BKodos for right answer..
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Re: The remainder when N is divided by 18 is 16. Given that N is [#permalink]
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18 Aug 2017, 04:17
Not sure about this big formule!!! but I got as per below logic.
We are been asked to find ... rem(N/(4*18))....
rem(n/18)=16... so hence (rem(n(4*18))=rem (16/4)=0... We can 0 remider only with two cases.. either 0 or 18.. so answer B is correct choice.



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Re: The remainder when N is divided by 18 is 16. Given that N is [#permalink]
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18 Aug 2017, 04:21
pratik1709 wrote: Not sure about this big formule!!! but I got as per below logic.
We are been asked to find ... rem(N/(4*18))....
rem(n/18)=16... so hence (rem(n(4*18))=rem (16/4)=0... We can 0 remider only with two cases.. either 0 or 18.. so answer B is correct choice. ............................... I guess your remainder values are wrong, see my solution and the remainder will be 13 and 4 which will come alternate as the N is increased in series./ Kudos for right answer
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The remainder when N is divided by 18 is 16. Given that N is [#permalink]
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18 Aug 2017, 18:40
smyarga wrote: 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?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5 N=28x N=18y+16 28x=18y+16➡ 7x4.5y=4 x=7 y=10 least value of N=28*7=196 N/4=196/4=49 49/18 gives a remainder of 13 LCM of 18 and 28=4*7*9=252 196+252=448=next value of N N/4=112 112/18 leaves a remainder of 4 (N/4)/18 leaves two cycling remainders, 13 and 4 2 B



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Re: The remainder when N is divided by 18 is 16. Given that N is [#permalink]
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19 Aug 2017, 06:30
Is this a GMAT question?



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Re: The remainder when N is divided by 18 is 16. Given that N is [#permalink]
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19 Aug 2017, 09:20
smyarga wrote: 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? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 Just nice problem from http://www.mualphatheta.org/National_Co ... Tests.aspxI know some ways how to solve it quickly. May be someone knows nicer way of solution. Thanks:) I am not sure of my method. Experts do let me know. Anyways, N= 18p+ 16 R Since N is multiple of 28, make it multiple. Divide by 4 (so its N/4 and divide by 18) N = (18*28p + 16*28R)/(4*18) N= 7p + 56/9R. Only 9 and 18 will make full remainder R. So, 2 values.



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Re: The remainder when N is divided by 18 is 16. Given that N is [#permalink]
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20 Aug 2017, 11:29
gps5441 wrote: Is this a GMAT question? Yes , it can be asked above or around 700 level difficulty level.
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Re: The remainder when N is divided by 18 is 16. Given that N is [#permalink]
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20 Sep 2017, 06:52
gracie wrote: smyarga wrote: 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?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5 N=28x N=18y+16 28x=18y+16➡ 7x4.5y=4 x=7 y=10 least value of N=28*7=196 N/4=196/4=49 49/18 gives a remainder of 13 LCM of 18 and 28=4*7*9=252 196+252=448=next value of N N/4=112 112/18 leaves a remainder of 4 (N/4)/18 leaves two cycling remainders, 13 and 4 2 B Hey, can you please explain the cycling remainders bit? I got the answer till ' remainder as 4'. Therefore I selected 1 as the answer. can you please explain?



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Re: The remainder when N is divided by 18 is 16. Given that N is [#permalink]
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20 Sep 2017, 07:24
akshay94raja wrote: gracie wrote: smyarga wrote: 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?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5 N=28x N=18y+16 28x=18y+16➡ 7x4.5y=4 x=7 y=10 least value of N=28*7=196 N/4=196/4=49 49/18 gives a remainder of 13 LCM of 18 and 28=4*7*9=252 196+252=448=next value of N N/4=112 112/18 leaves a remainder of 4 (N/4)/18 leaves two cycling remainders, 13 and 4 2 B Hey, can you please explain the cycling remainders bit? I got the answer till ' remainder as 4'. Therefore I selected 1 as the answer. can you please explain? HOPE IT HELPS 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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The remainder when N is divided by 18 is 16. Given that N is [#permalink]
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20 Sep 2017, 07:50
X, I be integers,
28X = 18I + 16
28X/4 => (18I+16)/4 => (9/2)I + 4
here (9/2)I leaves the remainder
For all even values of I,
(9/2)I leaves 0 as remainder
For all odd values of I.
(9/2) I leaves 1 as remainder
so we have 2 different remainders between 0 and 18 inclusive




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