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14 Dec 2023, 07:00
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D
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Difficulty:
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Question Stats:
18% (02:33) correct
82%
(02:57)
wrong
based on 449
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12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun
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
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
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16 Dec 2023, 13:23
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GMAT Club Official Explanation:
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.
General Discussion
14 Dec 2023, 07:26
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The possible cubes are +-1 +-8 +-27 +-64
Let's assume k=10a+b for a and b belonging to {0,1,2,......9}
k-f(k)=cube
9(a-b)=+-27 (cube can be negative)
Others are not divisible by 9 so mod(a-b)=3 is what we need to solve with a condition that a is not zero since such a case is forbidden.
(3,0),(4,1),(5,2),(6,3),(7,4),(8,5),(9,6),(0,3),(1,4),(2,5),(3,6),(4,7),(5,8),(6,9) are the solution of mod(a-b)=3 we reject (0,3) solution since that means k is single digit so 3 option(B)
Let's assume k=10a+b for a and b belonging to {0,1,2,......9}
k-f(k)=cube
9(a-b)=+-27 (cube can be negative)
Others are not divisible by 9 so mod(a-b)=3 is what we need to solve with a condition that a is not zero since such a case is forbidden.
(3,0),(4,1),(5,2),(6,3),(7,4),(8,5),(9,6),(0,3),(1,4),(2,5),(3,6),(4,7),(5,8),(6,9) are the solution of mod(a-b)=3 we reject (0,3) solution since that means k is single digit so 3 option(B)
