2007 AMC 8 Problems/Problem 18

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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 $B$?

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

Solution

$303\times505=153015$

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

$30303\times50505=1530453015$

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


See Also

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