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26 Apr 2013, 03:28
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B
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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
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Just nice problem from https://www.mualphatheta.org/National_C ... Tests.aspx
27 Apr 2013, 02:10
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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.
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.
madhurvashisht
Joined: 23 Jan 2013
Last visit: 10 May 2015
Posts: 31
Given Kudos: 12
Status:Application process
Location: India
Schools: AGSM '16 WBS '15 Copenhagen Cranfield '14 Erasmus HEC '15 Queen's '15 (D) Sauder '16 Schulich '15 (A) Broad '16
GMAT 1: 600 Q50 V22
GMAT 2: 650 Q49 V28
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WE:Information Technology (Computer Software)
Schools: AGSM '16 WBS '15 Copenhagen Cranfield '14 Erasmus HEC '15 Queen's '15 (D) Sauder '16 Schulich '15 (A) Broad '16
GMAT 2: 650 Q49 V28
Posts: 31
26 Apr 2013, 05:49
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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 .
\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 .
