Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 13"

Latest revision as of 10:33, 27 April 2008

The sum of the digits of the number $10^{2006}-2006$ is

$\mathrm{(A)}\ 18006\qquad\mathrm{(B)}\ 20060\qquad\mathrm{(C)}\ 2006\qquad\mathrm{(D)}\ 18047\qquad\mathrm{(E)}\ \text{None of these}$

$10^{2006}$ is a $1$ followed by 2006 $0$ 's. When we subtract $2006$ , we will get something close to 2006 $9$ 's.

The last four digits are $10000 - 2006 = 7994$ , and so we have 2002 $9$ s followed by $7994$ .

The sum of these is $2002 \cdot 9 + 7 + 9 + 9 + 4 = 18047 \Longrightarrow \mathrm{(D)}$

@@ Line 1: / Line 1: @@
 ==Problem==
-{{empty}}
+The sum of the digits of the number <math>10^{2006}-2006</math> is
+<math>\mathrm{(A)}\ 18006\qquad\mathrm{(B)}\ 20060\qquad\mathrm{(C)}\ 2006\qquad\mathrm{(D)}\ 18047\qquad\mathrm{(E)}\ \text{None of these}</math>
 ==Solution==
-{{solution}}
+<math>10^{2006}</math> is a <math>1</math> followed by 2006 <math>0</math>'s. When we subtract <math>2006</math>, we will get something close to 2006 <math>9</math>'s.
+The last four digits are <math>10000 - 2006 = 7994</math>, and so we have 2002 <math>9</math>s followed by <math>7994</math>.
+The sum of these is <math>2002 \cdot 9 + 7 + 9 + 9 + 4 = 18047 \Longrightarrow \mathrm{(D)}</math>
 ==See also==
 {{CYMO box|year=2006|l=Lyceum|num-b=12|num-a=14}}
+[[Category:Introductory Algebra Problems]]