12 Days of Christmas GMAT Competition - Day 4: For each positive

Question

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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

Bunuel · Accepted Answer

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.

Schinchan_bh · Answer

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)

Kinshook · Answer

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

$!vakumar.m · Answer

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

Kinshook · Answer

Bunuel  Please check the solution and correct OA if needed.

Bunuel · Answer

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.

Bunuel · Answer

Similar but a little bit easier question is here: https://gmatclub.com/forum/for-each-pos ... 21045.html

Adit_ · Answer

Point to note here is that 3^3 is the only absolute cube number possible

This is because every number and its reciprocal is of the form 10x+y and 10y+x
Subtracting:
9x-9y = 9(x-y)

In the list of cube numbers we only have 27 as our value.

We get 27 as our difference when x-y = 3.
So values from 14,25...and its reverse as well because negative numbers also possible.
There will be 13 values here.

9 more values being the big trick giving us 0

13+9 =  22 .

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