If x and y are positive integers, what is the value of x?

You are absolutely correct. I was incorrectly reading my target question as "What is the value of y?"

Hi Brent,
Great Q...
But statement 2 is sufficient as we are looking for value of x and that is 5..

Why is the OA not B.

Here is what i did for this question =>

We need the value of x.

Statement 1 => 9x=2xy+5
This tells us that x must be odd.
Cool.Lets use some test cases.

x=1,y=2 => Works
x=5,y=4 => Works

Hence not sufficient.
Statement 2 => xy=5y
Hence either y=0 or x=5
We are told that x,y are positive integers.
Hence x must be 5

Sufficient.

Smash that B.

Arggh!

I wrote that question a couple of weeks ago, and had the correct answer is D.
This morning, before posting it, I answered the question TWICE and got A both times.
Of course, I was incorrectly reading the target question as "What is the value of y"

I've changed the correct answer to D

Cheers and thanks,
Brent

Hi

In a hurry I missed the fact that y and x should be positive.

If y>0 then x - 5 should be = 0 and x = 5.

And 5y/5 will give us remainder 0 with any y.

Statement 2 is sufficient.

Unswer needs to be D.

Regards +1

nice question. is there any other simple way to solve this problem?
Thank you...

Target question: What is the value of x?

Statement 1: When 9x is divided by 2y, the quotient is x, and the remainder is 5
There's a nice rule that say, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3

So, from the above rule, we can write: 9x = 2yx + 5
Rewrite this as: 9x - 2yx = 5
Factor out the x to get: x(9 - 2y) = 5
Since x and 9 - 2y MUST BE INTEGERS, and since their PRODUCT is 5, we know that EITHER x = 1 and 9 - 2y = 5 OR x = 5 and 9 - 2y = 1

If x = 1 and 9 - 2y = 5, then x = 1 and y = 2
If x = 5 and 9 - 2y = 1, then x = 5 and y = 4

At this point it LOOKS LIKE there are two possible values for x. However, when we test these values, we have a problem.

If we plug x = 1 and y = 2 into statement 1, we get: When 9(1) is divided by 2(2), the quotient is 2, and the remainder is 1 (but we need a remainder of 5). So, x = 1 and y = 2 is NOT A SOLUTION

If we plug x = 5 and y = 4 into statement 1, we get: When 9(5) is divided by 2(4), the quotient is 5, and the remainder is 5. In other words, When 45 is divided by 8, the quotient is 5, and the remainder is 5. This WORKS. So, x = 5 and y = 4 IS A SOLUTION

So, we can conclude that x = 5
Since we can answer the target question with certainty, statement 1 is SUFFICIENT


Statement 2: When 5y is divided by x, the quotient is y, and the remainder is 0
So, from the above rule, we can write: 5y = xy + 0
Rewrite this as: 5y - xy = 0
Factor: y(5 - x) = 0
So, either y = 0, or x = 5
Since we're told y is POSITIVE, we know that y does not equal 0, which means x must equal 5
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer:

RELATED VIDEO

Yes, Thanks for video.

Hi Brent,

Is my understanding correct that x=1 and y=2 is being eliminated since the remainder 5 is greater than the divisor 2y? Or it violates the rule \(0<=r<d\)

Warm Regards,
Pritishd

Hi Pritishd,

You're correct.
The remainder (5) must always be less than the divisor (4).

I realize that I explained this part of my solution poorly. I've edited that part as follows:
If we plug x = 1 and y = 2 into statement 1, we get: When 9(1) is divided by 2(2), the quotient is 2, and the remainder is 1 (but we need a remainder of 5). So, x = 1 and y = 2 is NOT A SOLUTION

Statement 1 :
9x=2yx+5

2y > 5 --> 2y = 6,8,10,12,etc since 2y should be even

Substituting 2y values in the above equation :

2y = 6 --> 9x = 6x+5 --> 3x=5 Not possible since X has to be an integer
2y=8 --> 9x=8x+5 --> x=5
Sufficient

Statement 2:

5y=xy+0
5y=xy
x=5
Sufficient

Answer is D

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