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

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==Problem==
 
==Problem==
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The sum of the digits of the number <math>10^{2006}-2006</math> is
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A. <math>18006</math>
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B. <math>20060</math>
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C. <math>2006</math>
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D. <math>18047</math>
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E. None of these
  
 
==Solution==
 
==Solution==
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<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==
 
==See also==
 
{{CYMO box|year=2006|l=Lyceum|num-b=12|num-a=14}}
 
{{CYMO box|year=2006|l=Lyceum|num-b=12|num-a=14}}
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[[Category:Introductory Algebra Problems]]

Revision as of 19:23, 17 October 2007

Problem

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

A. $18006$

B. $20060$

C. $2006$

D. $18047$

E. None of these

Solution

$10^{2006}$ is a $1$ followed by 2006 $0$s; when we subtract, we will get something approximately 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 $2002 \cdot 9 + 7 + 9 + 9 + 4 = 18047 \Longrightarrow \mathrm{(D)}$

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 12
Followed by
Problem 14
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