1968 IMO Problems/Problem 2
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[hide]Problem
Find all natural numbers such that the product of their digits (in decimal notation) is equal to .
Video Solution (in Chinese)
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 |