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30 Dec 2023, 23:46
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D
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86%
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For each positive integer \(k\), \(f(k)\) is defined to be the number created by reversing the digits of \(k\). For instance, \(f(52) = 25, \ f(40) = 4\) and \(f(99) = 99\). How many positive two-digit integers \(k\) are there where \(k - f(k)\) is the cube of an integer?
A. 7
B. 13
C. 16
D. 22
E. 23
A. 7
B. 13
C. 16
D. 22
E. 23
30 Dec 2023, 23:46
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Official Solution:
For each positive integer \(k\), \(f(k)\) is defined to be the number created by reversing the digits of \(k\). For instance, \(f(52) = 25, \ f(40) = 4\) and \(f(99) = 99\). How many positive two-digit integers \(k\) are there where \(k - f(k)\) is the cube of an integer?
A. 7
B. 13
C. 16
D. 22
E. 23
Let's represent a two-digit number \(k\) as \(10a + b\), where \(a\) is the tens digit and \(b\) is the units digit. Then \(f(k)\) becomes \(10b + a\), and \(k - f(k) = (10a + b) - (10b + a) = 9(a - b) = 3^2(a - b)\).
For \(3^2(a - b)\) to be the cube of an integer, \((a - b)\) must be:
• 3, making \(k - f(k)\) equal to \(3^3 = 27\).
• -3, making \(k - f(k)\) equal to \(-3^3 = -27\).
• 0, making \(k - f(k)\) equal to \(0^3 = 0\).
Moving on to specific cases:
• \(a - b = 3\) is possible for the following 7 sets of \((a, b)\): (9, 6), (8, 5), (7, 4), (6, 3), (5, 2), (4, 1), and (3, 0).
• \(a - b = -3\) is possible for the following 6 sets of \((a, b)\): (1, 4), (2, 5), (3, 6), (4, 7), (5, 8), and (6, 9);
• \(a - b = 0\) is possible for the following 9 sets of \((a, b)\): (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8) and (9, 9).
Therefore, there are a total of 7 + 6 + 9 = 22 two-digit integers \(k\) where \(k - f(k)\) is the cube of an integer.
Answer: D
For each positive integer \(k\), \(f(k)\) is defined to be the number created by reversing the digits of \(k\). For instance, \(f(52) = 25, \ f(40) = 4\) and \(f(99) = 99\). How many positive two-digit integers \(k\) are there where \(k - f(k)\) is the cube of an integer?
A. 7
B. 13
C. 16
D. 22
E. 23
Let's represent a two-digit number \(k\) as \(10a + b\), where \(a\) is the tens digit and \(b\) is the units digit. Then \(f(k)\) becomes \(10b + a\), and \(k - f(k) = (10a + b) - (10b + a) = 9(a - b) = 3^2(a - b)\).
For \(3^2(a - b)\) to be the cube of an integer, \((a - b)\) must be:
• 3, making \(k - f(k)\) equal to \(3^3 = 27\).
• -3, making \(k - f(k)\) equal to \(-3^3 = -27\).
• 0, making \(k - f(k)\) equal to \(0^3 = 0\).
Moving on to specific cases:
• \(a - b = 3\) is possible for the following 7 sets of \((a, b)\): (9, 6), (8, 5), (7, 4), (6, 3), (5, 2), (4, 1), and (3, 0).
• \(a - b = -3\) is possible for the following 6 sets of \((a, b)\): (1, 4), (2, 5), (3, 6), (4, 7), (5, 8), and (6, 9);
• \(a - b = 0\) is possible for the following 9 sets of \((a, b)\): (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8) and (9, 9).
Therefore, there are a total of 7 + 6 + 9 = 22 two-digit integers \(k\) where \(k - f(k)\) is the cube of an integer.
Answer: D
26 Jan 2024, 11:48
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Similar but a little bit easier question is here: https://gmatclub.com/forum/m40-424807.html
