Question

From a two digit number N, we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then:

• A. N cannot end in 5
• B. N can end in any digit other than 5
• C. N does not exist
• D. there are exactly 7 values for N
• E. there are exactly 10 values for N

Answer 1

The correct option is D there are exactly 7 values for N

Let the two digit number be 10a+b

N = 10a+b

N' = reversed number = 10b+a

N-N' = 9(a-b)

N-N' is positive perfect cube.

$\therefore$ a>b

For 9(a-b) be perfect cube, a-b = 3

$\therefore$ b $ϵ$ [0, 6] $⟶$ 7 values

$\therefore$ a $ϵ$ [3, 9] $⟶$ 7 values

$\therefore$ Total 7 values are possible for N.