The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. I have been writing about some of the problems from the senior paper from 2018. A list of all of the problems can be found here.
Today we will look at the fifth problem from the 2018 South African Mathematics Olympiad:
Determine all sequences
of nonnegative integers such that
, and
divides
for all
.
Since the sequence is strictly increasing, we know that
for all positive integers
. (We could prove this rigorously by induction.) This means that
for all
, and so we know that
is equal to either
, or to
for all positive integers
. Perhaps we should try to figure out exactly when it is equal to
, and when it is equal to
. If we knew, for example, that we always have that
, then we have reduced the problem to solving this recurrence relation.
So when is it possible that ? Looking at our derivation above, we see that
is always strictly smaller than
unless both
and
. i.e. We must have equality in each of the two inequalities that we used.
We see that if there is some positive integer such that
, then we must have that
, and so, since the sequence is strictly increasing, we have that
for all
. The sequence thus has some initial segment where
, and then possibly switches to a sequence where
. In fact, it is possible that this switch never occurs. We could have that
for all positive integers
. This is in fact a solution to the problem. We verify that if
for all positive integers
, then
for all
.
We now turn our attention to finding all of the possible sequences where either there is no positive integer such that
, or where there is some positive integer
such that
, but
for all
.
Suppose that there are no positive integers such that
. We then have that
for all positive integers
. This implies that
for all positive integers . We recall, or learn, that the sum of the first
positive integers is
, and so we find that
for all positive integers . To be more rigorous, we could prove this using the principle of mathematical induction. By construction, this satisfies
for all
, and so this is a valid solution to the problem for all values of
.
Finally, what sequences arise if there is some positive integer such that
, but
for all
? Since
and the sequence is strictly increasing, we have that
for all positive integers
. We thus have that
for all
, and so we find using a similar argument to before that
for all positive integers .
Thus we potentially have a solution where for some positive integer , we have that
for
, and
for
. The conditions of the problem are clearly satisfied for
, since for these values of
we have that
, and by construction, we have that
for all
. Thus the only requirement that needs to be satisfied for this to be a valid solution is that
divides
. We thus require that
divides
. Since we have that
, we must in fact have that
. (We have that
, and so the multiple of
that
is equal to must be
itself.)
In summary, we found that all sequences that satisfy the conditions in the problem are given by
for all positive integers
; or
for all positive integers
, where the value of
can be chosen arbitrarily; or
- There is some positive integer
such that
for
, and
for all
.
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