2000 AIME II Problems/Problem 14
Contents
Problem
Every positive integer has a unique factorial base expansion , meaning that , where each is an integer, , and . Given that is the factorial base expansion of , find the value of .
Solution
Solution 1
Note that
Thus for all ,
So now,
Therefore we have , if for some , and for all other .
Therefore we have:
Solution 2 (informal)
This is equivalent to Solution 1. I put up this solution merely for learners to see the intuition.
Let us consider a base number system. It’s a well known fact that when we take the difference of two integral powers of , (such as ) the result will be an integer in base composed only of the digits and (in this example, ). More specifically, the difference , , is an integer digits long (note that has digits). This integer is made up of ’s followed by ’s.
It should make sense that this fact carries over to the factorial base, albeit with a modification. Whereas in the general base , the largest digit value is , in the factorial base, the largest digit value is the argument of the factorial in that place. (for example, is a valid factorial base number, as is . However, is not, as is greater than the argument of the second place factorial, . should be represented as , and is .) Therefore, for example, is not , but rather is . Thus, we may add or subtract factorials quite easily by converting each factorial to its factorial base expression, with a in the argument of the factorial’s place and ’s everywhere else, and then using a standard carry/borrow system accounting for the place value.
With general intuition about the factorial base system out of the way, we may tackle the problem. We use the associative property of addition to regroup the terms as follows: we now apply our intuition from paragraph 2. is equivalent to followed by ’s in the factorial base, and is followed by ’s, and so on. Therefore, followed by ’s in the factorial base. followed by ’s, and so on for the rest of the terms, except , which will merely have a in the place followed by ’s. To add these numbers, no carrying will be necessary, because there is only one non-zero value for each place value in the sum. Therefore, the factorial base place value is for all if , , and for all other .
Therefore, to answer, we notice that , and this will continue. Therefore, . We have 62 sets that sum like this, and each contains pairs of elements that sum to , so our answer is almost . However, we must subtract the in the place, and our answer is .
Solution 3 (less formality)
Let . Note that since (or is significantly smaller than ), it follows that . Hence . Then , and as , it follows that . Hence , and we now need to find the factorial base expansion of
Since , we can repeat the above argument recursively to yield , and so forth down to . Now , so .
The remaining sum is now just . We can repeatedly apply the argument from the previous two paragraphs to find that , and if for some , and for all other .
Now for each , we have . Thus, our answer is .
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.