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 2: / Line 2: @@
 The sum of the digits of the number <math>10^{2006}-2006</math> is
-A. <math>18006</math>
+<math>\mathrm{(A)}\ 18006\qquad\mathrm{(B)}\ 20060\qquad\mathrm{(C)}\ 2006\qquad\mathrm{(D)}\ 18047\qquad\mathrm{(E)}\ \text{None of these}</math>
-B. <math>20060</math>
+==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.
-C. <math>2006</math>
-D. <math>18047</math>
-E. None of these
+The last four digits are <math>10000 - 2006 = 7994</math>, and so we have 2002 <math>9</math>s followed by <math>7994</math>.
-==Solution==
+The sum of these is <math>2002 \cdot 9 + 7 + 9 + 9 + 4 = 18047 \Longrightarrow \mathrm{(D)}</math>
-<math>10^{2006}</math> is a <math>1</math> followed by 2006 <math>0</math>s; when we subtract, we will get something approximately 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==