If x and y are positive integers, what is the remainder when xy is div

For positive integers, yes. There is one more integer solution, though x = y = 0.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x and y are positive integers, what is the remainder when xy is divided by 4?

(1) x + y = xy

(2) (x – 1)(y – 1) is odd.

There are 2 variables and 2 equations are given from the 2 conditions, so there is high chance (C) will be our answer.
From condition, we only get x=y=2 as an answer, because xy-x-y=0, (x-1)(y-1)=1, x-1=1, y-1=1즉 x=y=2. This is sufficient.
From condition 2, in order for (x-1)(y-1)=odd, x-1=odd, y-1=odd, and x=1+odd=even, y=1+odd=even. Therefore, xy=even*even=4m(a multiple of 4), and the remainder when this is divided by 4 always becomes 0, so this is sufficient.
The answer becomes (D) according to common mistake type 4(B).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

(1 SUFFICIENT. One way to handle this statement is to consider even and odd numbers. Let's examine what kinds of numbers can satisfy the statement x + y = xy:

- If x and y are both odd, the left side is even, but the right side is odd. Therefore, this case is not possible.

- If one of x and y is odd and the other is even, the left side is odd, but the right side is even. This case, then, is impossible as well.

Only the case in which x and y are both even is possible. Therefore, both x and y are even, so each is a multiple of 2. It follows that the product xy is a multiple of 4, and so the desired remainder must be 0.

2) SUFFICIENT. If ( x – 1)( y – 1) is odd, then each of x – 1 and y – 1 must be odd. Therefore, both x and y are even, so each is a multiple of 2.

I might not be thinking this through. Is using smart numbers a wrong approach?

1) x + y = xy. The only numbers that satisfy this equation are 2 + 2 = 4, 1 + 4 = 4, and 1 + 3 = 3 (xy). However, the remainder for 2 + 2 = 4 is 0 and for 1 + 3 = 3 is 3 when divided by 4. How is it sufficient?

2) For the second one, I get it (x - 1)(y - 1) is odd, IF we assume x and y to be 2 and 2, it would be (2 - 1) ( 2 -1) which is 1 and odd. Hence this is sufficent.

Here, it's best to test cases for each statement.

In statement (1), we know that x+y = xy. If x and y are both odd, then x+y is even yet x*y is odd. So this case is not possible.

If x+y are both even, then x+y is even as is x*y.

Lastly, if x and y have different signs (order does not matter), then x+y is odd while x*y is even. The two sides of the equal sign do not match, so this case is not possible.

As a result, the only case that works occurs when both x and y are even. If x and y are both even, the their product is divisible by four since the product of two even numbers has at least two 2's in its prime factors.

Sufficient.

In statement (2), odd*odd = odd, so x and y are each one greater than an odd number. This fact means that x and y are even. The product of two even number is even. Sufficient.

there's an easier approach for statement 1:
y=x/(x-1);
since x,y are both positive int, x has to be 2. two consecutive no. are not divisible by each other except for 1&2.

If x and y are positive integers, what is the remainder when xy is divided by 4?

(1) x + y = xy

(2) (x – 1)(y – 1) is odd.

___________________________________________________________

from the statement 1,
xy - x - y = 0
adding +1 each side of hands,
1+xy-x-y = 1
--> (1-x)(1-y) = 1
since x and y are positive integers, the only way the above equation to be valid is x = y= 2
xy = 4, therefore the remainder is 0

from the statement 2,
because (x-1)(y-1) should be odd, both should be odd.
the only way both (x-1) and (y-1) to be odd is when x and y are even (even - 1(odd) = odd)
therefore, even * even is always divisible by 4.
sufficient.

the answer is D.

is a positive integer >0. so how can x and y=0 if both are positive integers?

Yes, positive integers are integers that are greater than 0. x = y = 0 is not possible for this question. My response there is just to demonstrate that generally x + y = xy over integers has two solution sets: x = y = 0 and x = y = 2.

Hi,

1) x+y=xy, the only integer that satisfies this equation is 2 ==> 2+2 = 2*2=4.
4/4 yields no remainder, Sufficient

2) (x – 1)(y – 1) is odd.
Develop : xy-x-y+1 = x(y-1)-(y-1)

We know that in a subtraction, to have a odd result you need Even-Odd or Odd-Even
if (y-1) is even, then x(y-1) must be too therefore it can't be Odd-Even

We are left with Even-Odd: it tells us that x is even and as y-1 is odd, y is even.

Therefore xy is divisible by 4 and will yield no remainder. Sufficient

Answer D

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