Proof. By Theorem 1, it suffices to show that any regular Kaprekar constant in each case is of the form stated in the assertion. In the following, let be any integer.
For any
b-adic five-digit regular Kaprekar constant
x, we denote by
with:
the rearrangement in descending order of the numbers of all digits of
x. By Ref. [
1] (Theorem 1.1 (7)),
We see the following magnitude relations among the numbers of all digits of
x:
Then, we obtain the following:
Proof. Since
is the maximum number among all digits of
x,
Since
is the second smallest number among all digits of
x, we then see that:
This implies that:
by the two inequalities above. Moreover, we see that:
which implies that:
as desired. □
We then see that the following equality holds:
This implies that
and:
Putting
, we then see that:
If , then we see a contradiction that is not regular. Therefore, , and Part is proven.
For any
b-adic seven-digit regular Kaprekar constant
x, we denote by
with:
the rearrangement in descending order of the numbers of all digits of
x. By the same argument as in the proof of Part
, we then see that one of the following two equalities holds:
The equality (i) implies a contradiction that .
The equality (ii) implies that
and:
Putting
, we then see that:
If , then we see a contradiction that is not regular. Therefore, , and Part is proven.
For any
b-adic nine-digit regular Kaprekar constant
x, we denote by
with:
the rearrangement in descending order of the numbers of all digits of
x. By the same argument as in the proof of Part
, we then see that one of the following six equalities holds:
The equalities (i) and (v) imply a contradiction that .
The equalities (iii), (iv), and (vi) imply a contradiction that .
The equality (ii) implies that
and:
Putting
, we then see that
x is equal to:
where the base
b is in the range
, since:
If , then we see a contradiction that b is in the range . Therefore, , and Part is proven.
For any
b-adic 11-digit regular Kaprekar constant
x, we denote by
with:
the rearrangement in descending order of the numbers of all digits of
x. By the same argument as in the proof of Part
, we then see that one of the following twenty equalities holds:
The equality (i) implies a contradiction that .
The equalities (ii), (x), (xii), and (xiii) imply a contradiction that .
The equalities (iii), (iv), (vii), (xi), and (xviii) imply a contradiction that .
The equality (v) implies a contradiction that .
The equalities (viii) and (xvi) imply a contradiction that .
The equality (ix) implies that:
which yields a contradiction that
.
The equality (xiv) implies a contradiction that .
The equality (xv) implies a contradiction that .
The equality (xvii) implies that:
which implies a contradiction that
.
The equality (xix) implies a contradiction that .
The equality (xx) implies a contradiction that .
The equality (vi) implies that
and:
Putting
, we then see that:
If , then we see a contradiction that is not regular. Therefore, , and Part is proven. □
Note that we shall need more calculations of solving simultaneous equations in the proof for even cases in Theorem 4 than odd cases in Theorem 3, because, in the case where is even, the Kaprekar transformation may not necessarily give us the maximum number among the numbers of all digits.
Proof. For any
b-adic two-digit regular Kaprekar constant
x, we denote by
with
the rearrangement in descending order of numbers of all digits of
x. By Ref. [
1] (Theorem 1.1 (2)),
We then see that one of the following two equalities holds:
The equality (i) implies a contradiction that .
The equality (ii) implies that:
Putting
, we then see that:
as desired.
For any
b-adic four-digit regular Kaprekar constant
x, we denote by
with
the rearrangement in descending order of the numbers of all digits of
x. By Ref. [
1] (Theorem 1.1 (6)),
Since:
we see that one of the following six equalities holds:
The equalities (i), (ii), and (vi) imply a contradiction that .
The equality (iii) implies that .
The equality (iv) implies a contradiction that .
The equality (v) implies that
and:
Putting
, we then see that:
If , then we see a contradiction that is not regular. Therefore, , and Part is proven.
For any
b-adic six-digit regular Kaprekar constant
x, we denote by
with:
the rearrangement in descending order of the numbers of all digits of
x. By Ref. [
1] (Theorem 1.1 (6)),
Since
and:
we see that
or
. The equality
implies a contradiction that
, and the equality
implies a contradiction that
. Therefore, we see that one of the following nine equalities holds:
The equality (i) implies that .
The equality (ii) and (iii) imply a contradiction that .
The equality (iv) implies that , which contradicts the condition that .
The equality (vi) implies a contradiction that .
The equality (vii) implies a contradiction that is not regular.
The equality (v) implies that
and:
Putting
with
, we then see that:
If , then we see a contradiction that is not regular. Therefore, .
The equality (viii) implies that
and:
Putting
, we then see that:
If , then we see a contradiction that is not regular. Therefore, .
The equality (ix) implies that
and:
Putting
, we then see that:
If , then we see a contradiction that is not regular. Therefore, , and Part is proven.
For any
b-adic eight-digit regular Kaprekar constant
x, we denote by
with:
the rearrangement in descending order of the numbers of all digits of
x. By Ref. [
1] (Theorem 1.1 (6)),
Since
and:
we see that
or
. The equality
implies a contradiction that
, and the equality
implies a contradiction that
. Therefore, we see that one of the following thirty equalities holds:
The equality (i) implies that .
The equality (ii) implies that .
The equality (iii) implies a contradiction that .
The equality (iv) implies a contradiction that .
The equalities (v), (x), (xv), and (xxi) imply a contradiction that .
The equality (vi) implies a contradiction that .
The equality (vii) implies a contradiction that .
The equalities (viii) and (ix) imply a contradiction that .
The equalities (xi), (xii), (xiii), and (xiv) imply a contradiction that .
The equality (xvii) implies a contradiction that .
The equality (xviii) implies a contradiction that .
The equality (xix) implies a contradiction that .
The equality (xx) implies a contradiction that .
The equality (xxii) implies a contradiction that .
The equality (xxiv) implies a contradiction that .
The equality (xxv) implies a contradiction that .
The equality (xxvi) implies a contradiction that .
The equality (xxvii) implies a contradiction that .
The equality (xxviii) implies a contradiction that .
The equality (xvi) implies that
and:
Putting
with
, we then see that:
If , then we see a contradiction that is not regular. Therefore, .
The equality (xxiii) implies that
and:
Putting
with
, we then see that:
If , then we see a contradiction that is not regular. Therefore, .
The equality (xxix) implies that
and:
Putting
, we then see that:
If , then we see a contradiction that is not regular. Therefore, .
The equality (xxx) implies that
and:
Putting
, we then see that:
If , then we see a contradiction that is not regular. Therefore, , and Part is proven. □