2007 AMC 8 Problems/Problem 18: Difference between revisions
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==Solution== | ==Solution== | ||
We can first make a small example to find out <math>A</math> and <math>B</math>. So, | |||
<math>303\times505=153015 </math> | <math>303\times505=153015 </math> | ||
The ones digit plus thousands digit is <math>5+3=8</math>. | The ones digit plus thousands digit is <math>5+3=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> | 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> | ||
Revision as of 19:57, 12 November 2017
Problem
The product of the two 
-digit numbers
and 
has thousands digit 
and units digit 
. What is the sum of and ?
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
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
