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The sum of two 3 digit numbers 3x5 and 6y3 is divisible by

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The sum of two 3 digit numbers 3x5 and 6y3 is divisible by [#permalink]  03 Jun 2008, 21:40
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The sum of two 3 digit numbers 3x5 and 6y3 is divisible by 9. If x > y then x =
x + y <10
The sum of 3x5 and 6y3 is a 3 digit numbers

I don’t agree with the OA. My point is where does A say that sum of 3x5 and 6y3 is a 3 digit number.
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Re: DS question [#permalink]  03 Jun 2008, 22:01
vdhawan1 wrote:
The sum of two 3 digit numbers 3x5 and 6y3 is divisible by 9. If x > y then x =
x + y <10
The sum of 3x5 and 6y3 is a 3 digit numbers

I don’t agree with the OA. My point is where does A say that sum of 3x5 and 6y3 is a 3 digit number.

I would go with D
1st option
6y3+3x5 should be divisible by 9
so, sum of all the digits should be divisible by 9, right?
6+y+3+3+x+5 =17+(x+y)
The number could be 18, 27 ,36 and so on
but the option says (x+y)<10
which means it can only be 18.
for this to happend x+y = 18-17 =1 right?
now, since x>y, x01, y=0
sufficient !

option B
If the sum of the two numbers is to be la 3 digit number,
there should be no carry fwd from (x+y), right?
hence (x+y)<10
sum of the digits+17+(x+y)
so x+1, y=0
sufficient

for me it is D. whats the OA please?
Re: DS question   [#permalink] 03 Jun 2008, 22:01
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The sum of two 3 digit numbers 3x5 and 6y3 is divisible by

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