Some remarks on Complements
Following Sudipta Das (see the Ed Pegg's page at
we define the n-complement of a number j as the number k
such that corresponding digits of j and k always add to n.
We want to investigate some properties of the n-complements in two special cases
for the terms j, k:
Let us sketch an idea that apply to the general problem (not only to our two cases).
Forbidding initial zeros on the representation of our numbers, we assume that
j, k have the same numbers of digits, say d; then we must have:
Main Formula: j+k=n*S where
S=111...1 is composed by d times the digit 1.
Of course carries could appear in the sum j+k, thus solutions of the new
problem could fail to solve the original one; but for the new formulation many
interesting properties will be quite immediate.
- the square-case, say
k=y2 for some x, y;
- the triangular-case, say
j=x*(x+1)/2, k=y*(y+1)/2 for some x, y.
Some trivial cases
Of course any choice of x, y between 0 and 3 will give rise to a solution
corresponding to a suitable n; thus we get trivial solutions for 10 values of
n = x2+ y2
say (choosing x £ y and sorting the results
with respect to n):
Of course we could do the same work concerning the two-digit-cases (say: x, y
both between 4 and 9), but it would be a quite tedious work, and a better idea could be
to ask the computer to do that. In fact, as we will show in a moment, we would waste our
time: no two-digits solution can exist.
A little bit of theory can save a lot of time...
Let us start from the "main formula" (in our case, say with
and remark that, once d has been fixed, we can easely find some factors of the
part S (the parts "thus n*S has the factor..." will be
explained in a moment):
- if d is even, the number S has the factor 11
(thus n*S has the factor 11*11 for n other than 11);
- if d is a multiple of 3, the number S has the factor
3 (thus n*S has the factor 3*3 if 3 does not divide
- if d is a multiple of 5, the number S has the factor
271 (thus n*S has the factor 271*271);
- if d is a multiple of 6, the number S has the factor
(thus n*S has the factor7*7 if 7 does not divide
- if d is a multiple of 7, the number S has the factor
239 (thus n*S has the factor 239*239);
- if d is a multiple of 13, the number S has the factor
79 (thus n*S has the factor 79*79);
Let us explain the special role played by the numbers 11, 3, 271, 7, 239, 79.
In general there is no relationship between the factors of x, y and the factors
of x2+ y2
(consider e.g. the example
74 = 2*37); however in some cases a very strong connection exists: special
factors of the sum must divide both x, y; so that also the square
of the factor divides the sum. This happens in particular for each one of the values
11, 3, 271, 7, 239, 79; of course, with respect to the representation given
in the main formula, some terms could divide n instead of the string of ones.