1968 IMO Problems/Problem 2
Problem
Find all natural numbers such that the product of their digits (in decimal notation) is equal to .
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,
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 |