Plug in numbers.
Pick C
Let N = 60,
N is a multiple of 20 -- YES
N+6 is a multiple of 3 - YES
Is the integer N a multiple of 15? --YES
Another example N = 120
N is a multiple of 20 -- YES
N+6 is a multiple of 3 - YES
Is the integer N a multiple of 15? --YES
St 1 & St2 alone is insufficient
Multiples
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Source: Beat The GMAT — Data Sufficiency |
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gmatmachoman
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from 1 N=20k (k=1,2,3....)
when N=60 then multiple of 15 but when N=80 the not a multiple of 15. So Not Suff.
from 2 N+6 = 3k (k=2,3,4....)
6 is a multiple of 3 so N will be also a multiple of 3 (Say 9=3+6 so N=3)
but we cannot say as some cases it will be multiple of 15 and some cases it will not be. So NS.
Combine both
if N=20k when k=1 then N=20 at the same time from 2 N+6 = 26 which will not be a multiple of 3
if N=20k when k=2 then N=40 at the same time from 2 N+6 = 46 which will not be a multiple of 3
if N=20k when k=3 then N=60 at the same time from 2 N+6 = 66 which will be a multiple of 3
if N=20k when k=4 then N=80 at the same time from 2 N+6 = 86 which will not be a multiple of 3
if N=20k when k=5 then N=100 at the same time from 2 N+6 = 106 which will not be a multiple of 3
if N=20k when k=6 then N=120 at the same time from 2 N+6 = 126 which will be a multiple of 3...
So when both (1) and (2) are satisfied we are getting a proper result.
IMOC
when N=60 then multiple of 15 but when N=80 the not a multiple of 15. So Not Suff.
from 2 N+6 = 3k (k=2,3,4....)
6 is a multiple of 3 so N will be also a multiple of 3 (Say 9=3+6 so N=3)
but we cannot say as some cases it will be multiple of 15 and some cases it will not be. So NS.
Combine both
if N=20k when k=1 then N=20 at the same time from 2 N+6 = 26 which will not be a multiple of 3
if N=20k when k=2 then N=40 at the same time from 2 N+6 = 46 which will not be a multiple of 3
if N=20k when k=3 then N=60 at the same time from 2 N+6 = 66 which will be a multiple of 3
if N=20k when k=4 then N=80 at the same time from 2 N+6 = 86 which will not be a multiple of 3
if N=20k when k=5 then N=100 at the same time from 2 N+6 = 106 which will not be a multiple of 3
if N=20k when k=6 then N=120 at the same time from 2 N+6 = 126 which will be a multiple of 3...
So when both (1) and (2) are satisfied we are getting a proper result.
IMOC
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Rahul@gurome
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(1) If N = 40, then N is not a multiple of 15
If N = 60, then N is a multiple of 15.
No unique answer.
So, (1) is NOT SUFFICIENT.
(2) If N = 30, then N + 6 = 36, is a multiple of 3 and N is a multiple of 15.
If N = 33, then N + 6 = 39, is a multiple of 3 and N is not a multiple of 15.
No unique answer.
So, (2) is NOT SUFFICIENT.
Combining (1) and (2), If N = 60, then N + 6 = 66, a multiple of 3; here N is a multiple of 15.
If N = 90, then N + 6 = 96, again a multiple of 3; here also N is a multiple of 15.
We can see for any other value also, that combining (1) and (2), we see that N is a multiple of 15.
[spoiler]The correct answer is (C).[/spoiler]
If N = 60, then N is a multiple of 15.
No unique answer.
So, (1) is NOT SUFFICIENT.
(2) If N = 30, then N + 6 = 36, is a multiple of 3 and N is a multiple of 15.
If N = 33, then N + 6 = 39, is a multiple of 3 and N is not a multiple of 15.
No unique answer.
So, (2) is NOT SUFFICIENT.
Combining (1) and (2), If N = 60, then N + 6 = 66, a multiple of 3; here N is a multiple of 15.
If N = 90, then N + 6 = 96, again a multiple of 3; here also N is a multiple of 15.
We can see for any other value also, that combining (1) and (2), we see that N is a multiple of 15.
[spoiler]The correct answer is (C).[/spoiler]
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N is a multiple of 15 if N is a multiple of both 3 and 5. From Statement 1, we learn that N is a multiple of 5. From Statement 2 we learn that N is a multiple of 3. So the answer is C.vongdn wrote:Is the integer N a multiple of 15?
(1) N is a multiple of 20
(2) N+6 is a multiple of 3
What is the method or process to solving this?
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