Difference between revisions of "2006 AMC 10B Problems/Problem 11"
m (→Solution) |
Dairyqueenxd (talk | contribs) (→Problem) |
||
| Line 2: | Line 2: | ||
What is the tens digit in the sum <math> 7!+8!+9!+...+2006!</math> | What is the tens digit in the sum <math> 7!+8!+9!+...+2006!</math> | ||
| − | <math> \ | + | <math> \textbf{(A) } 1\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 6\qquad \textbf{(E) } 9 </math> |
== Solution == | == Solution == | ||
Revision as of 12:58, 26 January 2022
Problem
What is the tens digit in the sum 
Solution
Since 
is divisible by 
, any factorial greater than is also divisible by . The last two digits of all factorials greater than are 
, so the last two digits of 
are .
(*)
So all that is needed is the tens digit of the sum 
So the tens digit is 
(*) A slightly faster method would be to take the 
residue of 
Since 
we can rewrite the sum as ![\[5040 + 8\cdot 5040 + 72\cdot 5040 \equiv 40 + 8\cdot 40 + 72\cdot 40 = 40 + 320 + 2880 \equiv 40 \pmod{100}.\]](http://latex.artofproblemsolving.com/1/d/e/1de56c07aa4efc6f508d9d71c952d4b824f74e02.png)
Since the last two digits of the sum is 
, the tens digit is 
See Also
| 2006 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
