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Re: 12 Days of Christmas GMAT Competition - Day 4: For each positive [#permalink]
Given: 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.
Asked: How many positive two-digit integers k are there where k - f(k) is the cube of an integer?

Let k = ab; where a is tenth digit and b is unit digit

k = 10a + b
f(k) = 10b + a
k - f(k) = (10a + b) - (10b + a) = 9(a-b)

Case 1:
a-b=3; k - f(k) = 27 = 3^3
(a, b) = {(3,0),(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)}:
Positive two-digit integers k are there where k - f(k) is the cube of an integer = {30,41,52,63,74,85,96} : 7 2-digit integers

Case 2:
a-b=-3; k - f(k) = -27 = (-3)^3
(a, b) = {(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)}
Positive two-digit integers k are there where k - f(k) is the cube of an integer = {14,25,36,47,58,69} : 6 2-digit integers

Total positive two-digit integers k are there where k - f(k) is the cube of an integer = 7 + 6 = 13

IMO B
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Re: 12 Days of Christmas GMAT Competition - Day 4: For each positive [#permalink]
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For each positive integer k, f(k) is defined to be the number created by reversing the digits of k.
How many positive two-digit integers k are there where k - f(k) is the cube of an integer?

let k be represented by xy -> 10x +y
then f(k) is yx -> 10y+x
then k -f(k) = 10x +y - 10y - x = 9(x-y) => the cube is a multiple of 9 or 3 (possible values are cubes of -729, -27, 0, 27, 729 etc )
For 9(x-y) = 0 => x =y
values of k are - 9 - 11, 22, 33, 44, 55, 66, 77, 88, 99

For 9(x-y) = - 27 => x-y = -3
or x = y-3 - possible values of k are 6 - 14, 25, 36, 47, 58, 69
For 9(x-y) = 27 => x-y = 3
or x = y+3 - possible values of k are 7 - 30, 41, 52, 63, 74, 85, 96

For 9(x-y) = -729 => x -y = -81
so we see that higher cube values are not possible as x and y are single digits

Total number of values of k are - 9 + 6+7 = 22
Answer D
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Re: 12 Days of Christmas GMAT Competition - Day 4: For each positive [#permalink]
Bunuel Please check the solution and correct OA if needed.

Kinshook wrote:
Given: 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.
Asked: How many positive two-digit integers k are there where k - f(k) is the cube of an integer?

Let k = ab; where a is tenth digit and b is unit digit

k = 10a + b
f(k) = 10b + a
k - f(k) = (10a + b) - (10b + a) = 9(a-b)

Case 1:
a-b=3; k - f(k) = 27 = 3^3
(a, b) = {(3,0),(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)}:
Positive two-digit integers k are there where k - f(k) is the cube of an integer = {30,41,52,63,74,85,96} : 7 2-digit integers

Case 2:
a-b=-3; k - f(k) = -27 = (-3)^3
(a, b) = {(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)}
Positive two-digit integers k are there where k - f(k) is the cube of an integer = {14,25,36,47,58,69} : 6 2-digit integers

Total positive two-digit integers k are there where k - f(k) is the cube of an integer = 7 + 6 = 13

IMO B
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Re: 12 Days of Christmas GMAT Competition - Day 4: For each positive [#permalink]
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Kinshook wrote:
Bunuel Please check the solution and correct OA if needed.

Kinshook wrote:
Given: 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.
Asked: How many positive two-digit integers k are there where k - f(k) is the cube of an integer?

Let k = ab; where a is tenth digit and b is unit digit

k = 10a + b
f(k) = 10b + a
k - f(k) = (10a + b) - (10b + a) = 9(a-b)

Case 1:
a-b=3; k - f(k) = 27 = 3^3
(a, b) = {(3,0),(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)}:
Positive two-digit integers k are there where k - f(k) is the cube of an integer = {30,41,52,63,74,85,96} : 7 2-digit integers

Case 2:
a-b=-3; k - f(k) = -27 = (-3)^3
(a, b) = {(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)}
Positive two-digit integers k are there where k - f(k) is the cube of an integer = {14,25,36,47,58,69} : 6 2-digit integers

Total positive two-digit integers k are there where k - f(k) is the cube of an integer = 7 + 6 = 13

IMO B


I think you are missing the following nine case when the digits of k are the same: 11, 22, 33, 44, 55, 66, 77, 88, and 99. In those cases k - f(k) = 0, which is a cube of an integer.
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Re: 12 Days of Christmas GMAT Competition - Day 4: For each positive [#permalink]
Expert Reply
Similar but a little bit easier question is here: https://gmatclub.com/forum/for-each-pos ... 21045.html
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Re: 12 Days of Christmas GMAT Competition - Day 4: For each positive [#permalink]
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