Difference between revisions of "2003 AMC 8 Problems/Problem 14"
m |
|||
| Line 13: | Line 13: | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2003|num-b=13|num-a=15}} | {{AMC8 box|year=2003|num-b=13|num-a=15}} | ||
| + | {{MAA Notice}} | ||
Revision as of 23:47, 4 July 2013
Problem
In this addition problem, each letter stands for a different digit.
If T = 7 and the letter O represents an even number, what is the only possible value for W?
Solution
Since both T's are 7, then O has to equal 4, because 7 + 7 = 14. Then, F has to equal 1. To get R, we do 4 + 4 (since O = 4) to get R = 8. The value for W then has to be a number less than 5, otherwise it will change the value of O, and can't be a number that has already been used, like 4 or 1. The only other possibilities are 2 and 3. 2 doesn't work because it makes U = 4, which is what O already equals. So, the only possible value of W is $\boxed{\text{\bolded{D}}}$ (Error compiling LaTeX. Unknown error_msg)
See Also
| 2003 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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. 
