Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 1"
m |
m |
||
| Line 9: | Line 9: | ||
The same argument applies to the units digit, and the sum of them is <math>45 \cdot 64 \cdot 1 = 2880</math>. Then <math>S = 324000+28800+2880 = 355\boxed{680}</math>. | The same argument applies to the units digit, and the sum of them is <math>45 \cdot 64 \cdot 1 = 2880</math>. Then <math>S = 324000+28800+2880 = 355\boxed{680}</math>. | ||
| − | == See | + | == See Also == |
{{Mock AIME box|year=Pre 2005|n=1|before=First question|num-a=2|source=14769}} | {{Mock AIME box|year=Pre 2005|n=1|before=First question|num-a=2|source=14769}} | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
Latest revision as of 12:44, 5 September 2012
Problem
Let 
denote the sum of all of the three digit positive integers with three distinct digits. Compute the remainder when is divided by 
.
Solution
We find the sum of all possible hundreds digits, then tens digits, then units digits. Every one of 
may appear as the hundreds digit, and there are 
choices for the tens and units digits. Thus the sum of the hundreds places is 
.
Every one of 
may appear as the tens digit; however, since 
does not contribute to this sum, we can ignore it. Then there are 
choices left for the hundreds digit, and choices afterwards for the units digit (since the units digit may also be ). Thus, the the sum of the tens digit gives 
.
The same argument applies to the units digit, and the sum of them is 
. Then 
.
See Also
| Mock AIME 1 Pre 2005 (Problems, Source) | ||
| Preceded by First question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
