Difference between revisions of "2009 AMC 10B Problems/Problem 1"
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== 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 10:10, 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.