Difference between revisions of "1968 IMO Problems/Problem 2"
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Find all natural numbers <math>x</math> such that the product of their digits (in decimal notation) is equal to <math>x^2 - 10x - 22</math>. | Find all natural numbers <math>x</math> such that the product of their digits (in decimal notation) is equal to <math>x^2 - 10x - 22</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/M2OfNZcUgSI | ||
==Solution 1== | ==Solution 1== |
Latest revision as of 19:18, 22 August 2024
Contents
Problem
Find all natural numbers such that the product of their digits (in decimal notation) is equal to
.
Video Solution
Solution 1
Let the decimal expansion of be
, where
are base-10 digits. We then have that
. However, the product of the digits of
is
, with equality only when
is a one-digit integer. Therefore the product of the digits of
is always at most
, with equality only when
is a base-10 digit. This implies that
, so
. Every natural number from 1 to 12 satisfies this inequality, so we only need to check these possibilities. It is easy to rule out 1 through 11, since
for those values. However,
, which is the product of the digits of 12. Therefore
is the only natural number with the desired properties.
Solution 2(SFFT)
It is pretty obvious that cannot be three digits or more, because then
is way too big.
Write where
and
are digits satisfying
. Then, we can use SFFT:
We have
It is therefore clear that
must be either
or
. We can then split into two cases:
We have or
, which is only satisfied when
or
.
We have . This is only satisfied when
, or
. Therefore,
, and so
~mathboy100
Solution 3
Let,
Now note that, if is a prime such that
then
.
That means,
But, which means
don't divivde
So, and
It is easy to see that has one solution and that is
( Prove it by contradiction)
So,
Remarks (added by pf02, August 2024)
Solutions 2 and 3 are not satisfactory. In fact, they can not be called solutions, since they make statements which are not proven. Specifically:
In Solution 2, the author writes "It is pretty obvious that
cannot be three digits or more, because then
is
way too big." This is intuitively true, but not obvious at all.
As a crucial step in the solution, it should be proven. Later,
the author states
".
It is therefore clear that
must be either
or
".
First, the last term should be instead of
. Either
way, the conclusion about
is not clear at all. As a second
crucial step in the solution, it should be proven.
In Solution 3, the notation and writing are very confusing. However, a diligent reader can make sense of them. But in this solution as well, there are statements which beg for a proof. The first such statement is
" which means
don't
divide
".
(When writing the author means the square of an
arbitrary natural number, not the square of the number
used in the line above this statement.) Neither the modulo
statements, nor the conclusion are obvious; proofs should
be given. The second unproven statement is
". It is easy to see that
has one
solution and that is
. (Prove it by contradiction.)"
(The author means "the equation has a unique solution for ".)
The conclusion about the uniqueness of
is not easy to see,
and as a crucial step in the solution, it should be proven.
Below, I will give corrected, complete, and somewhat simplified versions of these two solutions.
Solution 2 (corrected and complete)
Let the decimal expansion of be
,
where
are base-10 digits. Let us prove first that
.
Using and the fact that this expression
equals the product
we have that
.
Since
has
digits, we also have
. Thus, we have
. We will show that this can not be true
for
.
Divide by , rearrange factors and terms, and get
. Again using
, we get
.
This is false for
, so
the number of digits of
can not
be more than
.
Let then , with
the digits of
. We have
. By solving the quadratic equation
we see that the inequality is true when
.
At this point, we could simply check which of the numbers
has the product of its digits equal to
. It would turn out that
is the only one. But we can take a short cut by writing
. and using that
or
.
In the case
we get the equation
in
which has no integer solutions. In the case
we get the equation
in
, which has solutions
and
. Since
is a digit,
is the only acceptable solution, so
.
Solution 3 (corrected and complete)
Let . Since
is a product of digits, the only
prime factors of
can be
. Write
.
Note that if is a natural number then
or
.
Indeed,
or
or
for some
. By writing
out
the statement follows immediately. Then
or
, so
can not be a factor of
.
Similarly, if is a natural number then
or
.
Then
or
, so
can not be a factor
of
.
And finally if is a natural number then
or
.
Then
or
, so
can not be a
factor of
.
The only possible prime factor of is
. So
for some
. Write this as
, or
.
Clearly is not acceptable because the equation in
has no
integer solutions.
If the left hand side is divisible by
, but not by
. The right
hand side is not divisible by
if
is odd, and is divisible by
if
is even. So
yields no solutions.
If , the equation becomes
which has solutions
. The only acceptable solution is
.
See Also
1968 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |