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

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<math>\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10</math>
 
<math>\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10</math>
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==Video Solution==
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https://youtu.be/7an5wU9Q5hk?t=2085
  
 
==Solution==
 
==Solution==

Revision as of 20:18, 27 October 2020

Problem

The product of the two $99$-digit numbers

$303,030,303,...,030,303$ and $505,050,505,...,050,505$

has thousands digit $A$ and units digit $B$. What is the sum of $A$ and $B$?

$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10$

Video Solution

https://youtu.be/7an5wU9Q5hk?t=2085

Solution

We can first make a small example to find out $A$ and $B$. So,

$303\times505=153015$

The ones digit plus thousands digit is $5+3=8$.

Note that the ones and thousands digits are, added together, $8$. (and so on...) So the answer is $\boxed{\textbf{(D)}\ 8}$ This is a direct multlipication way.

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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 AJHSME/AMC 8 Problems and Solutions

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