2007 AMC 8 Problems/Problem 18: Difference between revisions
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/7an5wU9Q5hk?t=2085 | https://youtu.be/7an5wU9Q5hk?t=2085 | ||
==Solution== | ==Solution== | ||
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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 21:49, 20 April 2021
Problem
The product of the two 
-digit numbers
and 
has thousands digit 
and units digit 
. What is the sum of and ?
Video Solution
https://youtu.be/7an5wU9Q5hk?t=2085
Solution
We can first make a small example to find out and . So,
The ones digit plus thousands digit is 
.
Note that the ones and thousands digits are, added together, 
. (and so on...) So the answer is 
This is a direct multlipication way.
Video Solution by WhyMath
~savannahsolver
See Also
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
