Difference between revisions of "2000 AMC 8 Problems/Problem 20"
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==Problem== | ==Problem== | ||
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $<math>1.02</math>, with at least one coin of each type. How many dimes must you have?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $<math>1.02</math>, with at least one coin of each type. How many dimes must you have?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> | ||
| − | <math> \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5 </math> | + | <math>\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5</math> |
==Solution== | ==Solution== | ||
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Since you have one coin of each type, <math>1 + 5 + 10 + 25 = 41</math> cents are already determined, leaving you with a total of <math>102 - 41 = 61</math> cents remaining for <math>5</math> coins. | Since you have one coin of each type, <math>1 + 5 + 10 + 25 = 41</math> cents are already determined, leaving you with a total of <math>102 - 41 = 61</math> cents remaining for <math>5</math> coins. | ||
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==See Also== | ==See Also== | ||
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{{AMC8 box|year=2000|num-b=19|num-a=21}} | {{AMC8 box|year=2000|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:20, 11 February 2020
Problem
You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $
, with at least one coin of each type. How many dimes must you have?
Solution
Since you have one coin of each type, 
cents are already determined, leaving you with a total of 
cents remaining for 
coins.
You must have 
more penny. If you had more than penny, you must have at least 
pennies to leave a multiple of for the nickels, dimes, and quarters. But you only have more coins to assign.
Now you have 
cents remaining for 
coins, which may be nickels, quarters, or dimes. If you have only one more dime, that leaves 
cents in 
nickels or quarters, which is impossible. If you have two dimes, that leaves 
cents for 
nickels or quarters, which is again impossible. If you have three dimes, that leaves 
cents for nickel or quarter, which is still impossible. And all four remaining coins being dimes will not be enough.
Therefore, you must have no more dimes to assign, and the 
cents in coins must be divided between the quarters and nickels. We quickly see that nickels and quarters work. Thus, the total count is quarters, nickels, penny, plus one more coin of each type that we originally subtracted. Double-checking, that gives a total 
coins, and a total of 
cents.
There is only dime in that combo, so the answer is 
.
See Also
| 2000 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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. 
