x and y are positive integers satisfying 0

Question

x  and  y  are positive integers satisfying  0 < x - \sqrt{3}y < 1  and  b  is the decimal portion of  (x + \sqrt{3}y)^3 . What is the value of  (x - \sqrt{3}y)^3  in terms of  b ?

A.  b

B.  -b

C.  1 + b

D.  1 - b

E.  3b

MathRevolution · Accepted Answer

=>

Assume  a  is the decimal portion of  (x - \sqrt{3}y)^3.

Since the integer portion of  (x - \sqrt{3}y)^3  is  0 , we have  a = (x - \sqrt{3}y)^3.

If  c  is the integer portion of  (x + \sqrt{3}y)^3 , we have  (x + \sqrt{3}y)^3 = c + b.

Then, we have  (x - \sqrt{3}y)^3 + (x + \sqrt{3}y)^3 = 2(x^3 + 9xy^2) = a + c + b.

a + b = 2(x^3 + 9xy^2) - c

Since  x, y,  and  c  are positive integers,  a + b  is an integer.

Since we have  0 < a < 1  and  0 ≤ b < 1,  we have  0 < a + b < 2.

Thus  a + b = 1.

(x - \sqrt{3}y)^3 = a = 1 - b.

Therefore, D is the answer.
Answer: D

GraemeGmatPanda · Answer

Great question! Using the formulae: (c+d)^3 = c^3 + 3c^2d + 3cd^2 + d^3 and: (c-d)^3 = c^3 - 3c^2d + 3cd^2 - d^3 we have that: (c+d)^3 + (c-d)^3 = 2c^3 + 6cd^2 Applying this to c = x and d = \sqrt{3}y we have (x+\sqrt{3}y)^3 + (x-\sqrt{3}y)^3 = 2x^3 + 18xy^2 x and y are integers so 2x^3 + 18xy^2 is also an integer, and therefore has a decimal portion of 0 As a consequence the left hand side of the equation must also have a decimal portion of 0. ==> answer (D) 1-b

EDIT: My previous answer does not exclude (B): -b.

To eliminate (B) we can use the fact that: (x-\sqrt{3}y)^3 is positive since 0<(x-\sqrt{3}y)<1

ZIX · Answer

I used reasoning in this question.

If  (x - \sqrt{3}y)   is a +ve decimal number without any integer value (the expression is greater than 0 and less than 1), and x and y are integers (given) then:

Fact 1: Integer +/- Integer = integer 
Fact 2: Integer +/- (irrational number)*(Integer) = Integer with decimal value. This is what we are dealing with here.

Now  b  is the decimal value of  (x+ \sqrt{3}y) . So if the answer is 1.234 of this expression then 1 would come from integer x, and 0.234 would come from  \sqrt{3} y.

So if I have to express it in terms of b, it would be 1 + 0.234 or 1 +  b .

Similarly, for  (x - \sqrt{3}y)^3 . It will be expressed as 1 - b. Or decimal value  (here resulted by multiplication of   \sqrt{3}   with integer y)  is being subtracted from an integer value  (here represented by variable x) .  .

------------------------
PS: analyse it yourself. Don't just trust me

vallequalle · Answer

How can i avoid overlooking simple information like the "-"...?

x and y are positive integers satisfying 0 < x - 3y < 1 and b is the d

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions