Difference between revisions of "2007 AMC 8 Problems/Problem 18"

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When you multiply <math> 303 </math> by <math> 505 </math>, it's <math> 153015 </math>. The ones digit plus thousands digit is <math> 5 </math> <math> + </math> <math> 3 </math> <math> = </math> <math> 8 </math>. <math> 30303 </math> <math> \times </math> <math> 50505 </math> <math> = </math> <math> 1530453015 </math>. Note that the ones and thousands digits are, added together, <math> 8 </math>. (and so on...) So the answer is <math> \boxed{D} </math> <math> (8) </math>.
 
 
Answer is b
 
 
 
The question asks the person what the sum of the thousands digit is and the units digit from the second number,
 
 
 
therefore using 5+0=5, you should get the answer 5
 
 
 
 
 
 
 
Answer written by a third party person
 
Divyesh doddapaneni
 
--[[User:Divyesh|Divyesh]] 01:34, 1 November 2010 (UTC)
 

Revision as of 01:59, 14 November 2010

When you multiply $303$ by $505$, it's $153015$. The ones digit plus thousands digit is $5$ $+$ $3$ $=$ $8$. $30303$ $\times$ $50505$ $=$ $1530453015$. Note that the ones and thousands digits are, added together, $8$. (and so on...) So the answer is $\boxed{D}$ $(8)$.