Difference between revisions of "2000 AIME II Problems/Problem 14"
Mathgeek2006 (talk | contribs) m (→Solution 1) |
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So now, | So now, | ||
− | < | + | <cmath> |
\begin{align*} | \begin{align*} | ||
16!-32!+48!-64!+\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\cdots+(2000!-1984!)\\ | 16!-32!+48!-64!+\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\cdots+(2000!-1984!)\\ | ||
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&=16! +\sum_{m=1}^{62}\sum_{k=32m}^{32m+15}k\cdot k! | &=16! +\sum_{m=1}^{62}\sum_{k=32m}^{32m+15}k\cdot k! | ||
\end{align*} | \end{align*} | ||
− | </ | + | </cmath> |
Therefore we have <math>f_{16} = 1</math>, <math>f_k=k</math> if <math>32m\le k \le 32m+15</math> for some <math>m=1,2,\ldots,62</math>, and <math>f_k = 0</math> for all other <math>k</math>. | Therefore we have <math>f_{16} = 1</math>, <math>f_k=k</math> if <math>32m\le k \le 32m+15</math> for some <math>m=1,2,\ldots,62</math>, and <math>f_k = 0</math> for all other <math>k</math>. | ||
Line 25: | Line 25: | ||
Therefore we have: | Therefore we have: | ||
− | < | + | <cmath> |
\begin{align*} | \begin{align*} | ||
f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j &= (-1)^{17}\cdot 1 + \sum_{m=1}^{62}\sum_{k=32m}^{32m+15}(-1)^{k+1}k\\ | f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j &= (-1)^{17}\cdot 1 + \sum_{m=1}^{62}\sum_{k=32m}^{32m+15}(-1)^{k+1}k\\ | ||
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&= \boxed{495} | &= \boxed{495} | ||
\end{align*} | \end{align*} | ||
− | </ | + | </cmath> |
=== Solution 2 (less formality) === | === Solution 2 (less formality) === |
Revision as of 18:37, 13 March 2015
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 (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.