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 = 5Since 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 5Since we can answer the
target question with certainty, statement 2 is SUFFICIENT
Answer:
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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 \(0<=r<d\)