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X&Y are positive integers

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X&Y are positive integers

by debmalya_dutta » Tue Jul 13, 2010 7:15 pm
Need help with this . Any solution apart from pure number substitution

X&Y are positive integers, what is the remainder when x is divided by y?

1. when x is divided by 2y , the remainder is 4
2. when x+y is divided by y, the remainder is 4

Thanks in advance
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by Rahul@gurome » Tue Jul 13, 2010 8:47 pm
Consider statement (1) alone first.
Remainder is 4.
Remainder should always be less than the divisor.
So 4<2y
Or y >2. So y can be 3, 4, 5...
Let the quotient be "a" when x is divided by 2y.
So x = a*2y+4.
Or x/y = 2*a + 4/y.
Hence the remainder depends on the value 4/y.
If y is 3, the remainder is 1 . If y is 4 remainder is 0 and if y is more than 4 say 5, 6.. remainder is 4.
Since nothing definite can be said, (1) alone is not sufficient.

Next consider (2) alone.
Since remainder is less than the divisor, 4<y or y is 5, 6....
Let the quotient when (x+y) is divided by y be "b".
So x+y =b*y+4.
Or x = (b-1)*y+4.
Or x/y = (b-1) + 4/y.
Since y is more than 4 say 5, 6.. remainder is hence 4.
So statement (2) alone is sufficient to answer the question.

The correct answer is hence (B).
Last edited by Rahul@gurome on Tue Jul 13, 2010 10:31 pm, edited 1 time in total.
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by selango » Tue Jul 13, 2010 10:15 pm
@rahul

Remainder should always be less than quotient.

Let x=13,y=11

x/y;q=1,r=2

Here r>q

Let x=13,y=12

x/y;q=1;r=1

Here r=q

Remainder cannot be always less than quotient

Can you pl confirm...
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by Rahul@gurome » Tue Jul 13, 2010 10:32 pm
That was a mistake. It should be divisor. I have corrected in my earlier post.
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by heyabhi » Wed Jul 14, 2010 11:41 am
Great Explaination Rahul..
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