Difference between revisions of "1992 USAMO Problems/Problem 1"
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356412#356412 Discussion on AoPS/MathLinks] | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356412#356412 Discussion on AoPS/MathLinks] | ||
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
Latest revision as of 17:52, 27 April 2017
Problem
Find, as a function of 
the sum of the digits of
![\[9 \times 99 \times 9999 \times \cdots \times \left( 10^{2^n} - 1 \right),\]](http://latex.artofproblemsolving.com/3/9/5/395f3a29a3769af826b23ad1d2ca73fe825dd3d4.png)
where each factor has twice as many digits as the previous one.
Solution
The answer is 
.
Let us denote the quantity 
as 
. We wish to find the sum of the digits of .
We first note that
![\[P_{n-1} < \prod_{k=0}^{n-1} 10^{2^k} = 10^{2^n-1},\]](http://latex.artofproblemsolving.com/b/3/e/b3e9e84bdcf195f8c81e0bd3d062574dd20cdb37.png)
so 
is a number of at most 
digits. We also note that the units digit is not equal to zero. We may thus represent as
![\[\sum_{k=0}^{2^n-1} 10^k d_k ,\]](http://latex.artofproblemsolving.com/7/8/2/782da8e4923a29fa96c53eea46ce35e82e763dc2.png)
where the 
are digits and 
. Then
Thus the digits of are
![\[10-d_0, 9-d_1, 9-d_2, \dotsc, 9-d_{2^n-1}, d_0-1, d_1, d_2, \dotsc, d_{2^n-1} ,\]](http://latex.artofproblemsolving.com/0/6/2/062efb7a121862a1368e5b0075bf3a94bc51107b.png)
and the sum of these is evidently , as desired. 
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
| 1992 USAMO (Problems • Resources) | ||
| First Problem | Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
