Difference between revisions of "2009 AMC 10B Problems/Problem 1"
Whitelisted (talk | contribs) m (→Solution 2) |
Whitelisted (talk | contribs) m (→Problem) |
||
| Line 2: | Line 2: | ||
== Problem == | == Problem == | ||
| − | Each morning of her five-day workweek, Jane bought either a | + | Each morning of her five-day workweek, Jane bought either a <math>2000 fiji water or a </math>1000 espresso because she was smokin' rich. Her total cost for the week was a whole number of dollars, How many bottles of fiji water did she buy? |
| − | <math>\mathrm{(A)}\ | + | <math>\mathrm{(A)}\ 1000\qquad |
| − | \mathrm{(B)}\ | + | \mathrm{(B)}\ 2000\qquad |
| − | \mathrm{(C)}\ | + | \mathrm{(C)}\ 3000\qquad |
| − | \mathrm{(D)}\ | + | \mathrm{(D)}\ 4000\qquad |
| − | \mathrm{(E)}\ | + | \mathrm{(E)}\ 5000</math> |
== Solution 1 == | == Solution 1 == | ||
Revision as of 09:10, 21 May 2021
- The following problem is from both the 2009 AMC 10B #1 and 2009 AMC 12B #1, so both problems redirect to this page.
Contents
Problem
Each morning of her five-day workweek, Jane bought either a 
1000 espresso because she was smokin' rich. Her total cost for the week was a whole number of dollars, How many bottles of fiji water did she buy?
Solution 1
The only combination of five items with total cost a whole number of dollars is 3 muffins and 
bagels. The answer is 
.
Solution 2
Because 
ends in a 
, and we want a whole number of dollars, we know that there must be an even number of bagels. Furthermore, this tells us that the number of muffins is odd. Now, because it is a whole number of dollars, and 
cents multiplied by an odd number means that it will end in a , we know that the result of the even number multiplied by , must end in a . Note that the only result that gives this result is when is multiplied by 
. Thus, our answer is 
you dumb idiot.
~coolmathgames
See also
| 2009 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
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 AMC 10 Problems and Solutions | ||
| 2009 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by First Question |
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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
