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) .  .

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

vallequalle · Answer

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

voluptatibusnam · Answer

ig we can guess the ans to this is by: eliminating the options since (x-\sqrt{3}y) is btw 0 nd 1, implies cube of this will also be btw 0 nd 1 AND +ive . now c the options: A)b ..............ok, can be, but still doubtful as it cannot be same as (x+\sqrt{3})^3 B) -b.........clearly out, not +ve C) 1+b..........out, greater than 1 D)1-b..........ok, can be, also not similar to the (x+\sqrt{3})^3 .....cool! E) 3b...........can be greater than 1 whereas x nd y only follows 0

FreshStove · Answer

Can you please help me with the explanation and solution of this question? I am unable to comprehend it using the solutions/explanations provided.

SwethaReddyL · Answer

Bunuel  /  KarishmaB  - is this reliable approach?

KarishmaB · Answer

Yes, this is a good approximation though it can be made more accurate with very little extra work. In such questions, it is best to take a good guess and move on.

0 < x - \sqrt{3}y < 1

x and y are positive integers so say x = 2 and y = 1 to give us  2 - \sqrt{3} = 0.27  approx. Its cube is a very small value though it will stay positive. Cube of a positive is always positive. We need this cubed value in the options.

b is the decimal part of  (2 + \sqrt{3})^3 = (3.73)^3  approx.

Look at the options.

A.  b

The two decimals will not be the same because 1.73 is subtracted in one case and added in the other.

B.  -b

It cannot be negative as discussed above.

C.  1 + b

It is a very small value. It cannot be more than 1.

D.  1 - b

This is possible. This value will be less than 1.

E.  3b

As we said, it is a very small value so it is unlikely to be 3b.

Answer (D)

yukisuki · Answer

KarishmaB

Hi, how can we be sure the LHS has zero decimal portion, can't it be a.3 + b.7 = c.0 for example??

egmat · Answer

Hi yukisuki,

You're actually right that  in general  two numbers with decimals can add up to a whole number:  0.3  +  0.7  =  1.0  is a fine example. So it's a smart thing to double-check. The key is that here we're  not hoping  the decimals happen to line up - the algebra  forces  the sum to be an integer.

Look back at the step you're reacting to (Graeme's expansion). When you add the two cubes:

- (x + 3√y)3 contains "messy" cube-root terms,
- (x - 3√y)3 contains the  same  terms with opposite signs.

When you add them, those irrational pieces  cancel exactly , and what survives is built only from the integers x and y (Graeme wrote it as  2x3 + 18xy2 ). Any expression made purely by adding and multiplying integers  must  be a whole number - its decimal part is  0 , guaranteed, not by luck.

So the two expressions aren't two random decimals that happen to complement each other. They are  conjugates , and the structure of the subtraction/addition is what pins them together.

That's exactly why the final step works: if a is the decimal part of the small cube and b is the decimal part of the big cube, then a + b is an integer, and since both sit in  0 ,  1 ), the only value it can be is  1 .  Hence  a =  1  - b, giving  D .

Quick parallel to feel the "forced" cancellation:

- Compute ( 5  + √ 2 ) + ( 5  - √ 2 ).

The √ 2  and -√ 2  cancel, leaving exactly  10  - a whole number  every time , no coincidence required. Conjugate pairs always kill the irrational part. That's the same machinery running in this question.

Answer: D

KarishmaB · Answer

Yes, it can be. The solution does not claim that each term has 0 decimal part on the LHS. It claims that together the terms add up to give 0 decimal part and that makes sense, right? Because when they add up, we get  2x^3 + 18xy^2  which is an integer.

It means that if the two terms do have decimals, they add up to give 1.

As per your example, it means that if one term is a.3, then the other must be b.7 so that the decimal parts add up to give an integer. So if the decimal part of one number is b, the decimal part of the other will be 1 - b. The same is observed in your example too. Decimal part of one number if 0.3 and that of the other is 1 - 0.3 =0.7.

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

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