Difference between revisions of "2007 AMC 8 Problems/Problem 18"
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==Solution== | ==Solution== | ||
− | We can first make a small example to find out A and B So, | + | We can first make a small example to find out <math>A</math> and <math>B</math>. So, |
− | 303 | + | <math>303\times505=153015 </math> |
− | The ones digit plus thousands digit is 5+3=8 | + | The ones digit plus thousands digit is <math>5+3=8</math>. |
− | Note that the ones and thousands digits are, added together 8 (and so on...) So the answer is D | + | 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. | ||
Revision as of 11:22, 16 August 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.