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16 Apr 2020, 08:03
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
51% (02:38) correct
49%
(02:39)
wrong
based on 181
sessions
History
Date
Time
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Not Attempted Yet
\(x\) 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. \(3b\)
A. \(b\)
B. \(-b\)
C. \(1 + b\)
D. \(1 - b\)
E. \(3b\)
19 Apr 2020, 20:40
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=>
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
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
General Discussion
GraemeGmatPanda
Joined: 13 Apr 2020
Last visit: 21 Apr 2026
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16 Apr 2020, 08:27
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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 \(0<(x-\sqrt{3}y)<1\)
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 \(0<(x-\sqrt{3}y)<1\)
