2007 AMC 8 Problems/Problem 18

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 multiplication way.

Video Solution by WhyMath

https://youtu.be/_goaFuScO6M

~savannahsolver

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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2007 AMC 8 Problems/Problem 18

Contents

Problem

Video Solution

Solution

Video Solution by WhyMath

See Also