2007 AMC 8 Problems/Problem 18: Difference between revisions

Revision as of 22:47, 12 November 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 $\boxed{D}$ $(8)$ .

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 1: / Line 1: @@
+==Problem==
+The product of the two <math>99</math>-digit numbers
+<math>303,030,303,...,030,303</math> and <math>505,050,505,...,050,505</math>
+has thousands digit <math>A</math> and units digit <math>B</math>. What is the sum of <math>A</math> and <math>B</math>?
+<math>\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10</math>
+==Solution==
 <math>303\times505=153015 </math>
@@ Line 6: / Line 17: @@
 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>.
+==See Also==
+{{AMC8 box|year=2007|num-b=17|num-a=19}}