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

Revision as of 14:10, 26 December 2022

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 21: / Line 21: @@
 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>
-This is a direct multlipication way.
+This is a direct multiplication way.
 ==Video Solution by WhyMath==

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