2005 Indonesia MO Problems/Problem 2
Problem
For an arbitrary positive integer , define
as the product of the digits of
(in decimal). Find all positive integers
such that
.
Solution
First we will prove that must have 2 digits. Afterward, we'll let
and will solve for
.
Lemma: must have 2 digits
First, if , then
is a negative number, so
must have more than 1 digit.
Next, we will show that can not have 3 digits. Since 2005 is not a perfect square,
can not have any zeroes. That means the minimum value of
that has
digits is
. The maximum value of
with three digits is
. For the case where
, the minimum number is
and the maximum value of
is
. Since
, all three-digit values of
more than 111 have
, so
can not be a 3-digit number.
Finally, we will use induction to show that if can not have four or more digits. If
has
digits, where
, then the maximum value of
is
while the minimum value of
is
. For the base case, since
, we would have
for all four digit numbers, so no four digit number would work.
Assume that , where
. Multiplying both sides by 100 and adding 2005(99) results in
With the induction step complete, we've shown that
if
. Since
and
, we have
, so
can not be number with four or more digits. Thus,
must be a two-digit number.
From the Lemma, let . Substitution results in
Taking the equation modulo 4, we have
. That means
. Let
, resulting in
If
, then
, so
and
. Substitution results in
The only feasible value of
is
, so the only positive integer such that
is
. When plugging the value in, the equality holds.
See Also
2005 Indonesia MO (Problems) | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | Followed by Problem 3 |
All Indonesia MO Problems and Solutions |