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

Revision as of 12:24, 9 December 2012

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 $B$ ?

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

$303\times505=153015$

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 $\boxed{\textbf{(D)}\ 8}$

2007 AMC 8 (Problems • Answer Key • Resources)
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

@@ Line 16: / Line 16: @@
 <math> 30303\times50505=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>.
+Note that the ones and thousands digits are, added together, <math>8</math>. (and so on...) So the answer is <math>\boxed{\textbf{(D)}\ 8}</math>
 ==See Also==
 {{AMC8 box|year=2007|num-b=17|num-a=19}}