x and y are positive integers. When x is divided by y

When x is divided by y, the remainder is w : so w<y
When y is divided by x, the remainder is z : so z<x

i) x - y
say x=4 and y=2, w will be 0 and z will be 2....w+z = 0+2=2 and x-y=4-2=2...Possible

ii) x + y
we know w<y and z<x..Add the two equations w+z<x+y...Not possible that w+z=x+y

iii) y
say x=4 and y=2, w will be 0 and z will be 2....w+z = 0+2=2 and y=2...Possible

Thus i and iii

C
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We know that remainders are less than divisor and equal to or be more than 0.

Hence, 0<= w < y and 0<= z< x

Adding both,

0 <= w+z (required by stem) < x+y

Since its a could be question , lets try to prove them right. Lets back solve and substitute.

A) Can 0<= x-y < x+y ? YES -2y<0 or y>0 , which is a possible solution. So, there must be some numbers. Maybe true.

B) Can 0<= x+y < x+y ? No why ? Can 0<= 0 < 0 be possible ? No . So, can never be true.

C) Can 0<= y< x+y ? Yes 0<= 0 < x . Hence, inequality holds. There maybe some values that can satisfy the equations. Hence, can be true.


If one cannot analyze, plug simple numbers. Works !


Regards
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Given:
1. x and y are positive integers.
2. When x is divided by y, the remainder is w.
3. When y is divided by x, the remainder is z.

Asked: Which of the following COULD be the value of w+z?

x mod y = w ; x = k1y + w; If x>y; x = k1y + w; If x<y; x = w
y mod x = x; y = k2x + z; If x>y; y = z; if x<y; y = k2x + z

Case 1: x>y
w = x - k1y
z = y
w + z = x + y - k1y
When k1 = 2
w + z = x - y
When k1 = 0
w + z = x + y

Case 2: x<y
w = x
z = y - k2x
w + z = x + y - k2x
When k2 =1
w + z = y

IMO E

GMATPrepNow Bunuel

If it is not mentioned that x, y, w & z are different positive integers, some of them may be same, and the above solution is valid.
Please comment

That's correct.
If it is not mentioned that x, y, w & z are different positive integers, then some of them may be same.

Cheers,
Brent

for given infor
let
x=10
y=5
w=0
z=5
so w+z=5
since
10=5*2+0
5=10*0+5
so out of given options w+z=5 which is possible at
i) x - y & iii) y

option C

When x > y
x = k1 * y + w
In this case, k1 =/= 0 as if k1 has to be some number to suffice x > y.
So, in case 1, k1 = 0 is not possible. So, w + z = x + y is not a valid conclusion.

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