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Re: If x and y are positive integers, what is the value of x?
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27 Mar 2017, 10:17

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chetan2u wrote:

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

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"

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27 Mar 2017, 13:08

nice question. is there any other simple way to solve this problem? Thank you...
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“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ―Henry Wadsworth Longfellow

If x and y are positive integers, what is the value of x?
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Updated on: 06 Oct 2019, 07:43

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1

GMATPrepNow wrote:

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

1) When 9x is divided by 2y, the quotient is x, and the remainder is 5 2) When 5y is divided by x, the quotient is y, and the remainder is 0

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

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28 Mar 2017, 10:58

Yes, Thanks for video.
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“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ―Henry Wadsworth Longfellow

Re: If x and y are positive integers, what is the value of x?
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06 Oct 2019, 07:13

GMATPrepNow wrote:

GMATPrepNow wrote:

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

1) When 9x is divided by 2y, the quotient is x, and the remainder is 5 2) When 5y is divided by x, the quotient is y, and the remainder is 0

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 1, and the remainder is 5. We can see this this is NOT TRUE. 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

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 \(0<=r<d\)

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06 Oct 2019, 07:46

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Pritishd wrote:

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
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