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21 Aug 2014, 01:42
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Difficulty:
95%
(hard)
Question Stats:
44% (02:36) correct
56%
(01:55)
wrong
based on 91
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Is a three-digit number xyz divisible by 7?
(1) |2*z - 10x - y| is divisible by 7
(2) (z + 3y + 2x) is divisible by 7
(1) |2*z - 10x - y| is divisible by 7
(2) (z + 3y + 2x) is divisible by 7
Updated on: 06 Jul 2024, 03:29
Originally posted by BrushMyQuant on 21 Aug 2014, 07:28.
Last edited by BrushMyQuant on 06 Jul 2024, 03:29, edited 3 times in total.
Last edited by BrushMyQuant on 06 Jul 2024, 03:29, edited 3 times in total.
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Good Question!
Given 3 digit number is xyz and can be written as 100*x + 10*y + z
{ Ex: 123 can be written as 100*1 + 10*20 + 1*3 = 123 }
STATEMENT 1:
|2*z -10x -y| is divisible by 7
=> |2*z -10x - y| = 7k (where k is an integer)
=> 2z - 10x -y = 7r (where r can be positive or negative depending on the sign of 2z - 10x - y (doesn't really matter in this question till the time we know that its a multiple of 7))
Multiply both sides by -10 we get
-20z + 100x + 10y = -70r
-21z + z + 100x + 10y = -70r (breaking -20z into -21z + z)
or, 100x + 10y + z = 21z - 70r
21z - 70r is a multiple of 7
=> 100x + 10y + z is a multiple of 7, hence divisible by 7
So, xyz is divisible by 7 (xyz = 100*x + 10*y + 1*z)
So, SUFFICIENT
STATEMENT 2:
z + 3y + 2x = 7k (where k is an integer)
Multiply both sides by 50 we get
50z + 150y + 100x = 350k
49z + z + 140y+ 10y+ 100x = 350k
or, 100x + 10y + z = 350k - 140y - 49z
350k - 140y - 49z is divisible by 7
So, 100x + 10y + z is also divisible by 7
So, xyz is divisible by 7
So, SUFFICIENT
So, Answer will be D
Hope it helps!
Watch the following video to MASTER Absolute Values
Given 3 digit number is xyz and can be written as 100*x + 10*y + z
{ Ex: 123 can be written as 100*1 + 10*20 + 1*3 = 123 }
STATEMENT 1:
|2*z -10x -y| is divisible by 7
=> |2*z -10x - y| = 7k (where k is an integer)
=> 2z - 10x -y = 7r (where r can be positive or negative depending on the sign of 2z - 10x - y (doesn't really matter in this question till the time we know that its a multiple of 7))
Multiply both sides by -10 we get
-20z + 100x + 10y = -70r
-21z + z + 100x + 10y = -70r (breaking -20z into -21z + z)
or, 100x + 10y + z = 21z - 70r
21z - 70r is a multiple of 7
=> 100x + 10y + z is a multiple of 7, hence divisible by 7
So, xyz is divisible by 7 (xyz = 100*x + 10*y + 1*z)
So, SUFFICIENT
STATEMENT 2:
z + 3y + 2x = 7k (where k is an integer)
Multiply both sides by 50 we get
50z + 150y + 100x = 350k
49z + z + 140y+ 10y+ 100x = 350k
or, 100x + 10y + z = 350k - 140y - 49z
350k - 140y - 49z is divisible by 7
So, 100x + 10y + z is also divisible by 7
So, xyz is divisible by 7
So, SUFFICIENT
So, Answer will be D
Hope it helps!
Watch the following video to MASTER Absolute Values
21 Aug 2014, 07:35
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vad3tha
Nice Problem, Looks difficult but easy one.
First we need to check Is xyz is divisible by 7.
write 3 digit number as xyz= 100x+10y+z , You need to check is this number divisible.
Using Statement 1) (2z-10x-y) is divisible by 7
so it can be re written as 2z-10x-y = 7*N (where N is the interger which multiply 7 gives the result.)
10x = 2z-y-7n
putting the value of 10x in 100x+10y+z
10*10x+10y+z
10(2z-y-7n)+10y+z
20z-10y-70n+10y+z
21z-70N
7(3z-10N) So this is multiple of 7.
Using Statement 2) z + 3y + 2x = 7*K(where N is the interger which multiply 7 gives the result)
z=7k-3y-2x
putting this value of z in 100x+10y+z
100x+10y-7k-3y-2x
98x-7k+7y
7(14x-k+y)
which is also divisible by 7.
Hence D is the right answer.
