Difference between revisions of "2014 AMC 8 Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | We have <math>H=8-7=1</math> and <math>T=8-2+5=11</math>. Clearly <math>1-11= | + | We have <math>H=8-7=1</math> and <math>T=8-2+5=11</math>. Clearly <math>1-11=10</math> |
| − | , so our answer is <math>\boxed{\textbf{( | + | , so our answer is <math>\boxed{\textbf{(E)}10}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|before=First Problem|num-a=2}} | {{AMC8 box|year=2014|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:03, 11 January 2016
Problem
Harry and Terry are each told to calculate 
. Harry gets the correct answer. Terry ignores the parentheses and calculates 
. If Harry's answer is 
and Terry's answer is 
, what is 
?
Solution
We have 
and 
. Clearly 
, so our answer is 
.
See Also
| 2014 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
