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remainder

by ruplun » Fri May 06, 2011 9:07 pm
if w,x,y and z are the digits of the 4 digit number N ,a positive integer , what is the remeiander when N is divided by 9?

(1) w+x+y+z=13
(2)N+5 is divisible by 9
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by atulmangal » Fri May 06, 2011 11:14 pm
ruplun wrote:if w,x,y and z are the digits of the 4 digit number N ,a positive integer , what is the remeiander when N is divided by 9?

(1) w+x+y+z=13
(2)N+5 is divisible by 9
IMO D

(1) u can solve by putting some no's or if u know about the divisibility test of 9

for a no to be divisible by 9...the sum of digits should be a multiple of 9..for ex 72, 99, etc

try to see this pattern

72 --> 7+2 = 9....9/9 => R = 0 72/9 => R = 0
73 --> 7+3 = 10...10/9 => R = 1 73/9 => R = 1
74 --> 7+4 = 11...11/9 => R = 2 74/9 => R = 2

similarly for no N...the sum of digits is 13 always that means 13/9 => R = 4 will always be the remainder..

Op B...clearly 5 will always be the remainder

Hence Op D is correct
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by clock60 » Fri May 06, 2011 11:18 pm
hi ruplun
the answer looks like D
(1)w+x+y+z=13=9+4. remaider 4, it is possible to check simple numbers to verify
say ab=20, what remaider if divisible by 9
20=2*9+2. and 2+0=2=9*0+2
85=9*9+4, and 8+5=13=9*1+4
53=9*5+8, and 5+3=8=9*0+8
the same hols in the problem
the remaider will be the same as you divide the the sum of the digits by 9
so 1 st suff
(2)n+5 is divisible by 9
5 is not divisible by 9, and it comes that n is also not. but n+5 is divisible
thus the sum of remaiders resulting from division by 9 both 5 and must be divisible by 9
i mean
5=9*0+5, and the only remaider the if added to 5 will give sum that is divisible by 9 is 4
so 2 st also suff
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by clock60 » Fri May 06, 2011 11:20 pm
atulmangal wrote:
ruplun wrote:if w,x,y and z are the digits of the 4 digit number N ,a positive integer , what is the remeiander when N is divided by 9?

(1) w+x+y+z=13
(2)N+5 is divisible by 9
IMO D

(1) u can solve by putting some no's or if u know about the divisibility test of 9

for a no to be divisible by 9...the sum of digits should be a multiple of 9..for ex 72, 99, etc

try to see this pattern

72 --> 7+2 = 9....9/9 => R = 0 72/9 => R = 0
73 --> 7+3 = 10...10/9 => R = 1 73/9 => R = 1
74 --> 7+4 = 11...11/9 => R = 2 74/9 => R = 2

similarly for no N...the sum of digits is 13 always that means 13/9 => R = 4 will always be the remainder..

Op B...clearly 5 will always be the remainder
Hence Op D is correct
hi atul
you got right answer
but are you dead sure about st 2?
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by atulmangal » Fri May 06, 2011 11:32 pm
clock60 wrote:
atulmangal wrote:
ruplun wrote:if w,x,y and z are the digits of the 4 digit number N ,a positive integer , what is the remeiander when N is divided by 9?

(1) w+x+y+z=13
(2)N+5 is divisible by 9
IMO D

(1) u can solve by putting some no's or if u know about the divisibility test of 9

for a no to be divisible by 9...the sum of digits should be a multiple of 9..for ex 72, 99, etc

try to see this pattern

72 --> 7+2 = 9....9/9 => R = 0 72/9 => R = 0
73 --> 7+3 = 10...10/9 => R = 1 73/9 => R = 1
74 --> 7+4 = 11...11/9 => R = 2 74/9 => R = 2

similarly for no N...the sum of digits is 13 always that means 13/9 => R = 4 will always be the remainder..

Op B...clearly 5 will always be the remainder
Hence Op D is correct
hi atul
you got right answer
but are you dead sure about st 2?
Hey thanks for pointing that...remainder will be 4 always
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