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

(Solution)
Line 11: Line 11:
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/7an5wU9Q5hk?t=2085
 
https://youtu.be/7an5wU9Q5hk?t=2085
 +
  
 
==Solution==
 
==Solution==
Line 21: Line 22:
 
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>
 
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 multlipication way.
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/_goaFuScO6M
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2007|num-b=17|num-a=19}}
 
{{AMC8 box|year=2007|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:49, 20 April 2021

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 multlipication way.

Video Solution by WhyMath

https://youtu.be/_goaFuScO6M

~savannahsolver

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png